100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	Find a number, such that one-fourth of the number is 3 more than 7:1. 402. 603. 804. 100
Answer» Correct Answer - Option 1 : 40 Calculation: Let the number be x Now, A/Q x/4 = 7 + 3 ⇒ x = 10 × 4 ⇒ x = 40 ∴ The required number is 40

1.

Find a number, such that one-fourth of the number is 3 more than 7:1. 402. 603. 804. 100

Answer» Correct Answer - Option 1 : 40

Calculation:

Let the number be x

Now, A/Q

x/4 = 7 + 3

⇒ x = 10 × 4

⇒ x = 40

∴ The required number is 40

Discussion

2.	If ‘+’ stands for division, ‘×’ stands for addition, ‘–’ stands for multiplication, and ‘÷’ stands for subtraction, then which of the following equation is correct?(a) 36 × 6 + 7 ÷ 2 – 6 = 2(b) 36 + 6 – 3 × 5 ÷3 = 24(c) 36 ÷ 6 + 3 × 5 – 3 = 45(d) 36 – 6 + 3 × 5 ÷ 3 = 74
Answer» 36 × 6 ÷3 + 5 – 3 ⇒ 36 × 2 + 5 – 3 = 74

2.

If ‘+’ stands for division, ‘×’ stands for addition, ‘–’ stands for multiplication, and ‘÷’ stands for subtraction, then which of the following equation is correct?(a) 36 × 6 + 7 ÷ 2 – 6 = 2(b) 36 + 6 – 3 × 5 ÷3 = 24(c) 36 ÷ 6 + 3 × 5 – 3 = 45(d) 36 – 6 + 3 × 5 ÷ 3 = 74

Answer»

36 × 6 ÷3 + 5 – 3

⇒ 36 × 2 + 5 – 3 = 74

Discussion

3.	An 8 - digit number 4985x76y is divisible by 45. Find the difference between the minimum and maximum values of (3x + y).1. 112. 103. 154. 14
Answer» Correct Answer - Option 2 : 10 Given - 8 - digit number 4985x76y is divisible by 45 Concept used - For divisibility of 5, unit digit must be either 0 or 5 For divisibility of 9, sum of the digits divisible by 9 Solution - 8 - digit number 4985x76y If y = 0 the number divisible by 5 For divisibility of 9, ⇒ 4 + 9 + 8 + 5 + x + 7 + 6 + 0 = 39 + x, so x = 6 Value of (3x + y) = 3 × 6 + 0 = 18 If y = 5 the number divisible by 5. For divisibility of 9, ⇒ 4 + 9 + 8 + 5 + x + 7 + 6 + 5 = 44 + x, so, x = 1 Value of (3x + y) = 3 × 1 + 5 = 8 Difference between minimum and maximum value = (18 - 8) = 10. ∴ Difference between the minimum and maximum value = 10.

3.

An 8 - digit number 4985x76y is divisible by 45. Find the difference between the minimum and maximum values of (3x + y).1. 112. 103. 154. 14

Answer» Correct Answer - Option 2 : 10

Given -

8 - digit number 4985x76y is divisible by 45

Concept used -

For divisibility of 5, unit digit must be either 0 or 5

For divisibility of 9, sum of the digits divisible by 9

Solution -

8 - digit number 4985x76y

If y = 0 the number divisible by 5

For divisibility of 9,

⇒ 4 + 9 + 8 + 5 + x + 7 + 6 + 0 = 39 + x, so x = 6

Value of (3x + y) = 3 × 6 + 0 = 18

If y = 5 the number divisible by 5.

For divisibility of 9,

⇒ 4 + 9 + 8 + 5 + x + 7 + 6 + 5 = 44 + x, so, x = 1

Value of (3x + y) = 3 × 1 + 5 = 8

Difference between minimum and maximum value = (18 - 8) = 10.

∴ Difference between the minimum and maximum value = 10.

Discussion

4.	The total of the ages of Amar, Akbar and Anthony is 80 years. What was the total of their ages three years ago ?
Answer» Here, required sum = (80 – 3 x 3) years = (80 – 9) years = 71 years.

Discussion

5.	If the vectors \( -3 i+4 j+\lambda k \) and \( \mu i+8 j+6 k \) are collinear vectors, then find \( \lambda \) and \( \mu \).
Answer» Given that ⇒ the vectors -3i + 4j + λk and μi + 8j + 6k are collinear vectors. Collinear means the vectors lying on the Same Line. So the vectors can be Parallel Anti-parallel In both cases, the ratio of their respective components will be Equal. So we have -3i + 4j + λk & μi + 8j + 6k Comparing both, we get -3/μ = 4/8 = λ/6 -3/μ = 1/2 = λ/6 ⇒ -3/μ = 1/2 And 1/2 = λ/6 ⇒ μ/-3 = 2/1 And λ/6 = 1/2 ⇒ μ = -3(2) And λ = 6(1/2) ⇒ μ = - 6 And λ = 3

5.

If the vectors \( -3 i+4 j+\lambda k \) and \( \mu i+8 j+6 k \) are collinear vectors, then find \( \lambda \) and \( \mu \).

Answer»

Given that

⇒ the vectors -3i + 4j + λk and μi + 8j + 6k are collinear vectors.

Collinear means the vectors lying on the Same Line.

So the vectors can be

Parallel
Anti-parallel

In both cases, the ratio of their respective components will be Equal.

So we have

-3i + 4j + λk &

μi + 8j + 6k

Comparing both, we get

-3/μ = 4/8 = λ/6

-3/μ = 1/2 = λ/6

⇒ -3/μ = 1/2 And 1/2 = λ/6

⇒ μ/-3 = 2/1 And λ/6 = 1/2

⇒ μ = -3(2) And λ = 6(1/2)

⇒ μ = - 6 And λ = 3

Discussion

6.	The coordinates of three points A(-3/2,3), B(6,-2) and C(-3,4). Using vectors show that three points are collinear.
Answer» Given points are A(-3/2,3), B(6,-2) and C(-3,4) \(\vec{AB}\) = \(\vec{OB}\) - \(\vec{OA}\) = (6\(\hat i\) - 2\(\hat j\)) - (-3/2\(\hat i\) + 3\(\hat j\)) = (6 + 3/2)\(\hat i\) - (2+3)\(\hat j\) = 15/2 \(\hat i\) - 5\(\hat j\) \(\vec{BC}\) = \(\vec{OC}\) - \(\vec{OB}\) (-3\(\hat i\) + 4\(\hat j\)) - (6\(\hat i\) - 2\(\hat j\)) = (-3 - 6)\(\hat i\) + (4 + 2)\(\hat j\) = - 9\(\hat i\) + 6\(\hat j\) Now, \(\vec{AB}\) . \(\vec{BC}\) = (15/2\(\hat i\) - 5\(\hat j\)) . (-9\(\hat i\)+ 6\(\hat j\)) = 15/2 x -9 - 5 x 6 = \(\cfrac{-135}2\) - 30 = \(\cfrac{-135-60}2\) = \(\cfrac{-195}2\) \(\left\|\vec{AB}\right\|\) = \(\sqrt{(\cfrac{15}2)^2+(-5)^2}\) = \(\sqrt{\cfrac{225}4+25}\) = \(\sqrt{\cfrac{{225}+{100}}4}\) = \(\sqrt{\cfrac{325}2}\) = \(\cfrac{5\sqrt{13}}2\) \(\left\|\vec{BC}\right\|\) = \(\sqrt{(-9)^2+6^2}\) = \(\sqrt{81+36}\) = \(\sqrt{117}\) = \(3\sqrt{13}\) Cos θ = \(\cfrac{{\vec{AB}}.{\vec{BC}}}{{\|\vec{AB}\|}\|{\vec{BC}\|}}\) = \(\cfrac{\cfrac{-195}2}{\cfrac{5\sqrt{13}}{2}\times3\sqrt{13}}\) = - 1 = cos \(\pi\) Hence, angle between \(\vec {AB}\) and \(\vec {BC}\) is 180° \(\therefore\) Points A,B and C are collinear

6.

The coordinates of three points A(-3/2,3), B(6,-2) and C(-3,4). Using vectors show that three points are collinear.

Answer»

Given points are A(-3/2,3), B(6,-2) and C(-3,4)

\(\vec{AB}\) = \(\vec{OB}\) - \(\vec{OA}\)

= (6\(\hat i\) - 2\(\hat j\)) - (-3/2\(\hat i\) + 3\(\hat j\))

= (6 + 3/2)\(\hat i\) - (2+3)\(\hat j\)

= 15/2 \(\hat i\) - 5\(\hat j\)

\(\vec{BC}\) = \(\vec{OC}\) - \(\vec{OB}\)

(-3\(\hat i\) + 4\(\hat j\)) - (6\(\hat i\) - 2\(\hat j\))

= (-3 - 6)\(\hat i\) + (4 + 2)\(\hat j\)

= - 9\(\hat i\) + 6\(\hat j\)

Now, \(\vec{AB}\) . \(\vec{BC}\) = (15/2\(\hat i\) - 5\(\hat j\)) . (-9\(\hat i\)+ 6\(\hat j\))

= 15/2 x -9 - 5 x 6

= \(\cfrac{-135}2\) - 30

= \(\cfrac{-135-60}2\) = \(\cfrac{-195}2\)

\(\left|\vec{AB}\right|\) = \(\sqrt{(\cfrac{15}2)^2+(-5)^2}\)

= \(\sqrt{\cfrac{225}4+25}\)

= \(\sqrt{\cfrac{{225}+{100}}4}\) = \(\sqrt{\cfrac{325}2}\)

= \(\cfrac{5\sqrt{13}}2\)

\(\left|\vec{BC}\right|\) = \(\sqrt{(-9)^2+6^2}\)

= \(\sqrt{81+36}\) = \(\sqrt{117}\)

= \(3\sqrt{13}\)

Cos θ = \(\cfrac{{\vec{AB}}.{\vec{BC}}}{{|\vec{AB}|}|{\vec{BC}|}}\)

= \(\cfrac{\cfrac{-195}2}{\cfrac{5\sqrt{13}}{2}\times3\sqrt{13}}\) = - 1 = cos \(\pi\)

Hence, angle between \(\vec {AB}\) and \(\vec {BC}\) is 180°

\(\therefore\) Points A,B and C are collinear

Discussion

7.	A sells goods to B at a loss of 10%, B sells it to C at a loss of 10%, C sells it to D at a loss of 10%, if D pay Rs.3,645 for it, Find loss of B.1. Rs. 5502. Rs. 5003. Rs. 4004. Rs. 4505. None of these
Answer» Correct Answer - Option 4 : Rs. 450 Given: A’s loss percent = 10% B’s loss percent = 10% C’s loss percent = 10% D’s cost price = Rs.3,645 Concept: Loss = Cost price – Selling price Calculation: Let, cost price for A be Rs.1000x Cost price of B = {(100 – 10)/100} × 1000x = Rs.900x Cost price of C = {(100 – 10)/100} × 900x = Rs.810x Cost price of D = {(100 – 10)/100} × 810x = Rs.729x = 3645 ⇒ x = 5 Loss of B = (900x – 810x) × 5 = Rs.450 ∴ Loss of B is Rs.450.

7.

A sells goods to B at a loss of 10%, B sells it to C at a loss of 10%, C sells it to D at a loss of 10%, if D pay Rs.3,645 for it, Find loss of B.1. Rs. 5502. Rs. 5003. Rs. 4004. Rs. 4505. None of these

Answer» Correct Answer - Option 4 : Rs. 450

Given:

A’s loss percent = 10%

B’s loss percent = 10%

C’s loss percent = 10%

D’s cost price = Rs.3,645

Concept:

Loss = Cost price – Selling price

Calculation:

Let, cost price for A be Rs.1000x

Cost price of B = {(100 – 10)/100} × 1000x = Rs.900x

Cost price of C = {(100 – 10)/100} × 900x = Rs.810x

Cost price of D = {(100 – 10)/100} × 810x = Rs.729x = 3645

⇒ x = 5

Loss of B = (900x – 810x) × 5 = Rs.450

∴ Loss of B is Rs.450.

Discussion

8.	A sells goods to B at a profit of 25%, B sells it to C at a profit of 20%, C sells it to D at a loss of 30%, if D pay Rs.2,100 for it, Find the difference between cost price of B and C.1. Rs.4002. Rs.5503. Rs.4504. Rs.5005. None of these
Answer» Correct Answer - Option 4 : Rs.500 Given: A’s profit percent = 25% B’s profit percent = 20% C’s loss percent = 30% D’s cost price = Rs.2,100 Calculation: Let, cost price for A = Rs.100x Cost price of B = {(100 + 25)/100} × 100x = Rs.125x Cost price of C = {(100 + 20)/100} × 125x = Rs.150x Cost price of D = {(100 – 30)/100} × 150x = Rs.105x = 2100 ⇒ x = 20 Cost price of B = 125 × 20 = Rs.2,500 Cost price of C = 150 × 20 = Rs.3,000 Difference of cost price of B and C = 3000 – 2500 = Rs.500 ∴ Difference of cost price of B and C is Rs.500.

8.

A sells goods to B at a profit of 25%, B sells it to C at a profit of 20%, C sells it to D at a loss of 30%, if D pay Rs.2,100 for it, Find the difference between cost price of B and C.1. Rs.4002. Rs.5503. Rs.4504. Rs.5005. None of these

Answer» Correct Answer - Option 4 : Rs.500

Given:

A’s profit percent = 25%

B’s profit percent = 20%

C’s loss percent = 30%

D’s cost price = Rs.2,100

Calculation:

Let, cost price for A = Rs.100x

Cost price of B = {(100 + 25)/100} × 100x = Rs.125x

Cost price of C = {(100 + 20)/100} × 125x = Rs.150x

Cost price of D = {(100 – 30)/100} × 150x = Rs.105x = 2100

⇒ x = 20

Cost price of B = 125 × 20 = Rs.2,500

Cost price of C = 150 × 20 = Rs.3,000

Difference of cost price of B and C = 3000 – 2500 = Rs.500

∴ Difference of cost price of B and C is Rs.500.

Discussion

9.	A sells goods to B at a profit of 20%, B sells it to C at a profit of 50%, C sells it to D at a profit of 30%, if D pay Rs.936 for it, what was the cost price for B?1. Rs.5802. Rs.5003. Rs.4754. Rs.4505. None of these
Answer» Correct Answer - Option 5 : None of these Given: A’s profit percent = 20% B’s profit percent = 50% C’s profit percent = 30% D’s cost price = Rs.936 Calculation: Let, cost price for A = Rs.100x Cost price of B = {(100 + 20)/100} × 100x = Rs.120x Cost price of C = {(100 + 50)/100} × 120x = Rs.180x Cost price of D = {(100 + 30)/100} × 180x = Rs.234x = 936 ⇒ x = 4 Cost price of B = 120 × 4 = Rs.480 ∴ Cost price for B is Rs.480.

9.

A sells goods to B at a profit of 20%, B sells it to C at a profit of 50%, C sells it to D at a profit of 30%, if D pay Rs.936 for it, what was the cost price for B?1. Rs.5802. Rs.5003. Rs.4754. Rs.4505. None of these

Answer» Correct Answer - Option 5 : None of these

Given:

A’s profit percent = 20%

B’s profit percent = 50%

C’s profit percent = 30%

D’s cost price = Rs.936

Calculation:

Let, cost price for A = Rs.100x

Cost price of B = {(100 + 20)/100} × 100x = Rs.120x

Cost price of C = {(100 + 50)/100} × 120x = Rs.180x

Cost price of D = {(100 + 30)/100} × 180x = Rs.234x = 936

⇒ x = 4

Cost price of B = 120 × 4 = Rs.480

∴ Cost price for B is Rs.480.

Discussion

10.	Ranjan sells a Pendrive for Rs 910 at 30% profit and another for Rs 180 at 10% loss. His total profit or loss % is:A. 27.95%B. 24.50%C. 21.11%D. 22.11%1. C2. D3. B4. A
Answer» Correct Answer - Option 1 : C Given: Selling price = Rs. 910 And profit = 30% Another Selling price = Rs. 180 And loss = 10% Formula used: Profit = Cost price – Selling price Profit% = (Profit × 100)/Cost price Cost price = [{100/(100 + profit %)} × Selling price] Cost price = [{100/(100 – loss %)} × Selling price] Calculation: According to the question: ⇒ CP of first Pen drive = [{100/(100 + 30%)} × 910 ⇒ CP = [(100/130) × 910] ⇒ CP = Rs. 700 Again, ⇒ CP of second Pen drive = [{100/(100 – 10%)} × 180 ⇒ CP = [(100/90) × 180] ⇒ CP = Rs. 200 Total CP = (700 + 200) = Rs. 900 Total SP = (910 + 180) = Rs. 1090 Profit = (1090 – 900) = Rs. 190 Profit % = (190/900) × 100 ⇒ 19/9 = 21.11% ∴ Total profit is 21.11%.

10.

Ranjan sells a Pendrive for Rs 910 at 30% profit and another for Rs 180 at 10% loss. His total profit or loss % is:A. 27.95%B. 24.50%C. 21.11%D. 22.11%1. C2. D3. B4. A

Answer» Correct Answer - Option 1 : C

Given:

Selling price = Rs. 910

And profit = 30%

Another Selling price = Rs. 180

And loss = 10%

Formula used:

Profit = Cost price – Selling price

Profit% = (Profit × 100)/Cost price

Cost price = [{100/(100 + profit %)} × Selling price]

Cost price = [{100/(100 – loss %)} × Selling price]

Calculation:

According to the question:

⇒ CP of first Pen drive = [{100/(100 + 30%)} × 910

⇒ CP = [(100/130) × 910]

⇒ CP = Rs. 700

Again,

⇒ CP of second Pen drive = [{100/(100 – 10%)} × 180

⇒ CP = [(100/90) × 180]

⇒ CP = Rs. 200

Total CP = (700 + 200) = Rs. 900

Total SP = (910 + 180) = Rs. 1090

Profit = (1090 – 900) = Rs. 190

Profit % = (190/900) × 100

⇒ 19/9 = 21.11%

∴ Total profit is 21.11%.

Discussion

11.	6 years ago Rahul was 12 years old how old is he now
Answer» Given, 6 years ago Rahul was 12 years. Therefore his present age would be 12+6=18 years

Discussion

12.	Give the complete solution of this general linear homogeneous function.y'''- y'' = e3x + 3e4x
Answer» Given differential equation is y'''-y''=e^3x+3 e^4x It's auxiliary equation is m³-m² = 0 = m²(m-1) = 0 = m = 0, 0 & 1. ∴ CF = (C₁ + C₂x)e0x + C₃e^x = C₁ + C₂x + C₃e^x P.I. = \(\frac {1}{D^3-D^2} (e^{3x} + 3e^{4x)}\) = \(\frac {1}{D^2(D-1)}e^{3x} + \frac {3}{D^2(D-1)}e^{4x}\) = \(\frac {e^{3x}}{3^2(3-1)} + \frac {3e^{4x}}{4^2(4-1)}\) = \(\frac {e^{3x}}{18} + \frac {3e^{4x}}{48}\) = \(\frac {e^{3x}}{18} + \frac {e^{4x}}{16}\) ∴ Complete solution of given differential equation is y = CF + P.I. = C₁ + C₂x + C₃e^x + \(\frac {e^{3x}}{18} + \frac {e^{4x}}{16}\)

12.

Give the complete solution of this general linear homogeneous function.y'''- y'' = e3x + 3e4x

Answer»

Given differential equation is y'''-y''=e^3x+3 e^4x

It's auxiliary equation is

m³-m² = 0

= m²(m-1) = 0

= m = 0, 0 & 1.

∴ CF = (C₁ + C₂x)e0x + C₃e^x

= C₁ + C₂x + C₃e^x

P.I. = \(\frac {1}{D^3-D^2} (e^{3x} + 3e^{4x)}\)

= \(\frac {1}{D^2(D-1)}e^{3x} + \frac {3}{D^2(D-1)}e^{4x}\)

= \(\frac {e^{3x}}{3^2(3-1)} + \frac {3e^{4x}}{4^2(4-1)}\)

= \(\frac {e^{3x}}{18} + \frac {3e^{4x}}{48}\)

= \(\frac {e^{3x}}{18} + \frac {e^{4x}}{16}\)

∴ Complete solution of given differential equation is

y = CF + P.I.

= C₁ + C₂x + C₃e^x + \(\frac {e^{3x}}{18} + \frac {e^{4x}}{16}\)

Discussion

13.	In each of the following number series, the wrong number is given, find out that number.3, 4.5, 7, 7.5, 91. 32. 4.53. 7.54. 75. None of the above
Answer» Correct Answer - Option 4 : 7 Calculation∶ The series follows following pattern 0.5 + 1 + 1.5 = 3 1 + 1.5 + 2 = 4.5 1.5 + 2 + 2.5 = 6 2 + 2.5 + 3 = 7.5 2.5 + 3 + 3.5 = 9 ∴ The wrong number in the series is 7

13.

In each of the following number series, the wrong number is given, find out that number.3, 4.5, 7, 7.5, 91. 32. 4.53. 7.54. 75. None of the above

Answer» Correct Answer - Option 4 : 7

Calculation∶

The series follows following pattern

0.5 + 1 + 1.5 = 3

1 + 1.5 + 2 = 4.5

1.5 + 2 + 2.5 = 6

2 + 2.5 + 3 = 7.5

2.5 + 3 + 3.5 = 9

∴ The wrong number in the series is 7

Discussion

14.	In each of the following number series, the wrong number is given, find out that number.12, 12, 16, 34, 84, 1821. 122. 843. 344. 165. 182
Answer» Correct Answer - Option 2 : 84 Calculation∶ The series follows following pattern ⇒ 12 + (1³ – 1²) = 12 ⇒ 12 + (2³ – 2²) = 16 ⇒ 16 + (3³ – 3²) = 34 ⇒ 34 + (4³ – 4²) = 82 ⇒ 82 + (5³ – 5²) = 182 ∴ The wrong number in the series is 84

14.

In each of the following number series, the wrong number is given, find out that number.12, 12, 16, 34, 84, 1821. 122. 843. 344. 165. 182

Answer» Correct Answer - Option 2 : 84

Calculation∶

The series follows following pattern

⇒ 12 + (1³ – 1²) = 12

⇒ 12 + (2³ – 2²) = 16

⇒ 16 + (3³ – 3²) = 34

⇒ 34 + (4³ – 4²) = 82

⇒ 82 + (5³ – 5²) = 182

∴ The wrong number in the series is 84

Discussion

15.	If W/X = 6/11, Y/Z = 16/23 and W/Y = 9/16. Then what is the value of X/Y?1. 16/112. 33/323. 23/164. 32/33
Answer» Correct Answer - Option 2 : 33/32 Given: W/X = 6/11, Y/Z = 16/23 and W/Y = 9/16. Calculation: X/W = 11/6, Z/Y = 23/16, W/Y = 9/16 X/W × Z/Y × W/Y = XZ/(Y × Y) = X/Y × Z/Y ⇒ X/Y × 23/16 = 11/6 × 23/16 × 9/16 ⇒ X/Y = 99/96 = 33/32

15.

If W/X = 6/11, Y/Z = 16/23 and W/Y = 9/16. Then what is the value of X/Y?1. 16/112. 33/323. 23/164. 32/33

Answer» Correct Answer - Option 2 : 33/32

Given:

W/X = 6/11, Y/Z = 16/23 and W/Y = 9/16.

Calculation:

X/W = 11/6,

Z/Y = 23/16,

W/Y = 9/16

X/W × Z/Y × W/Y = XZ/(Y × Y) = X/Y × Z/Y

⇒ X/Y × 23/16 = 11/6 × 23/16 × 9/16

⇒ X/Y = 99/96 = 33/32

Discussion

16.	Simplify:(D2 + 4D + 5) * y = 0
Answer» Create separate equations (d² + 4d + 5)y = 0 (d² + 4d +5) = 0 y = 0 *(D² + 4D + 5) y = 0 Reorder the terms:** (5 + 4D + D²) * y = 0 Reorder the terms for easier multiplication: y(5 + 4D + D²) = 0 (5 * y + 4D * y + D² * y) = 0 (5y + 4yD + yD²) = 0 Solving: 5y + 4yD + yD² = 0 Solving for variable 'y'. Move all terms containing y to the left, all other terms to the right. Factor out the Greatest Common Factor (GCF), 'y'. y(5 + 4D + D²) = 0

16.

Simplify:(D2 + 4D + 5) * y = 0

Answer»

Create separate equations

(d² + 4d + 5)y = 0

(d² + 4d +5) = 0

y = 0

(D² + 4D + 5) * y = 0

Reorder the terms:

(5 + 4D + D²) * y = 0

Reorder the terms for easier multiplication:

y(5 + 4D + D²) = 0

(5 * y + 4D * y + D² * y) = 0

(5y + 4yD + yD²) = 0

Solving:

5y + 4yD + yD² = 0

Solving for variable 'y'.

Move all terms containing y to the left, all other terms to the right.

Factor out the Greatest Common Factor (GCF), 'y'.

y(5 + 4D + D²) = 0

Discussion

17.	The curve between charge density `rho` and distance r near `p - N` junction will beA. B. C. D.
Answer» Correct Answer - A Fact about `P - N` junction.

Discussion

18.	If a particle is seen through layers of parallel glass slabs of different refractive index as shown in figure. Where `mu_(0)` is the medium of the observer. If the object has velocity `v` towarsd observer at this instant along the line joining object and observer then, which of the following correctly represent the velocity of the object as seen by observer?A. `v_(app)=(V)/(m_(3)//mu_(0))`B. `v_(app)=(v)/(mu_(1)//mu_(0))`C. `v_(app)=(v)/(mu_(1)//mu_(0))+(v)/(mu_(2)//mu_(0))+(v)/(mu_(3)//mu_(0))`D. `v_(app)=v`
Answer» Correct Answer - A Apparent velocity `=((v)/(mu_(3)//mu_(2)))/((mu_(2)//mu_(1))/(mu_(1)//mu_(0)))=(v)/(mu_(3)//mu_(0))`

Discussion

19.	A carbohydrates which can not be digested in human gut is
Answer» Answer: Cellulose

Discussion

20.	A T.V. tower has a height 150m. What is the population density around the T.V. tower if the total population covered is 50 lakh? (Radius of earth `=6.4xx10^(6)m)`A. `708.5 km^(-2)`B. `828.6 km^(-2)`C. `856.8 km^(-2)`D. `454 km^(-2)`
Answer» Correct Answer - B Population covered `= rho pi d^(2)` `50 xx 10^(5) = rho (pi hR)` `:. Rho = (50 xx 10^(5))/(3.14 xx 2 xx 150 xx (6.4 xx 10^(6))) = 828.6 km^(-2)`

Discussion

21.	The cost of 2 rings and 4 bangles is Rs.46854. What is the cost of 5 rings and 10 bangles ?(a) Rs.115345 (b) Rs.117135(c) Rs.116675 (d) Cannot be determined(e) None of these
Answer» (b) Let the CP of 1 ring and 1 bangle be Rs.x and Rs.y respectively. 2x + 4y = 46854 2.5 (2x + 4y) = 2.5 × 46854 5x + 10y = Rs.117135

Discussion

22.	The difference between a two-digit number and the number obtained after interchanging the two digits of the two- digit number is 27. The sum of the two digits of the two-digit number is 15. What is the two-digit number ?(a) 87 (b) 96(c) 69 (d) Cannot be determined(e) None of these
Answer» (d) Let the two digit nubmer be 10x + y, where x is the first digit and y the second digit. (10x + y) – (10y + x) = 27 9x – 9y = 27 x – y = 3 ....(a) also x + y = 15 ....(b) x = 9 and y = 6 Required number is 96 or 69

Discussion

23.	If A = (1, 2, 3, 4) and B = (2, 4, 5), then (A - B) ∪ (B - A) =1. (1, 3, 5)2. (2, 4, 5)3. (2, 4)4. (3, 4, 5)
Answer» Correct Answer - Option 1 : (1, 3, 5) Given: A = (1, 2, 3, 4) B = (2, 4, 5) Concept used: If two sets X and Y, X - Y = Set of all the elements present in X but not in Y Y - X = Set of all the elements present in Y but not in X X ∪ Y = Set of all the elements present in either X or Y Calculation: A - B = (1, 2, 3, 4) - (2, 4, 5) ⇒ (1, 3) B - A = (2, 4, 5) - (1, 2, 3, 4) ⇒ (5) (A - B) ∪ (B - A) = (1, 3) ∪ (5) ⇒ (1, 3, 5) ∴ The value of (A - B) ∪ (B - A) is (1, 3, 5).

23.

If A = (1, 2, 3, 4) and B = (2, 4, 5), then (A - B) ∪ (B - A) =1. (1, 3, 5)2. (2, 4, 5)3. (2, 4)4. (3, 4, 5)

Answer» Correct Answer - Option 1 : (1, 3, 5)

Given:

A = (1, 2, 3, 4)

B = (2, 4, 5)

Concept used:

If two sets X and Y,

X - Y = Set of all the elements present in X but not in Y

Y - X = Set of all the elements present in Y but not in X

X ∪ Y = Set of all the elements present in either X or Y

Calculation:

A - B = (1, 2, 3, 4) - (2, 4, 5)

⇒ (1, 3)

B - A = (2, 4, 5) - (1, 2, 3, 4)

⇒ (5)

(A - B) ∪ (B - A) = (1, 3) ∪ (5)

⇒ (1, 3, 5)

∴ The value of (A - B) ∪ (B - A) is (1, 3, 5).

Discussion

24.	If the sum of four consecutive even numbers is 228, which is the smallest of the numbers ?(a) 52 (b) 54(c) 56 (d) 48(e) None of these
Answer» (b) According to the question, x + x + 2 + x + 4 + x + 6 = 228 4x + 12 = 228 4x = 228 – 12 = 216 x = 216/4 = 54 The smallest even number = 54

24.

If the sum of four consecutive even numbers is 228, which is the smallest of the numbers ?(a) 52 (b) 54(c) 56 (d) 48(e) None of these

Answer»

(b) According to the question,
x + x + 2 + x + 4 + x + 6 = 228
4x + 12 = 228

4x = 228 – 12 = 216
x = 216/4 = 54
The smallest even number = 54

Discussion

25.	Sulphuric acid is used to prepare more volatile acids from their corresponding salts due to its :(A) strong acidic nature(B) low volatility(C) strong affinity for water(D) ability to act as a dehydrating agent
Answer» Option : (B) low volatility Concentrated sulphuric acid has a high boiling point. So, it is considered to be non-volatile. Therefore, It is used for preparing volatile acids from their salts.

25.

Sulphuric acid is used to prepare more volatile acids from their corresponding salts due to its :(A) strong acidic nature(B) low volatility(C) strong affinity for water(D) ability to act as a dehydrating agent

Answer»

Option : (B) low volatility

Concentrated sulphuric acid has a high boiling point.

So, it is considered to be non-volatile.

Therefore,

It is used for preparing volatile acids from their salts.

Discussion

26.	An element with density 6 g cm-3 forms a fcc lattice with edge length of 4 x 10-8 cm. The molar mass of the element is (NA = 6 x 1023 mol-1) :(A) 57.6 g mol-1(B) 28.8 g mol-1(C) 82.6 g mol-1(D) 62 g mol-1
Answer» Option : (A) 57.6 g mol^-1 We know that, Number of atoms per unit cell of FCC = 4 Z = 4 We have given - d = 6g/cm³, a = edge length = 4 x 10^-8 cm N_A = 6 x 10²³ d = \(\frac{ZM}{a^3N_A}\) M = molar mass of element. ∴ 6g/cm³ = \(\frac{4\times M}{(4\times 10^{-8})^3cm^3\times 6\times 10^{23}mol^{-1}}\) ⇒ 6 x (64 x 10^-24) x 6 x 10²³ = 4M ⇒ M = \(\frac{6\times 6\times 64\times 10^{-1}}{4}g/mol\) ∴ M = 57.6 g/mol.

26.

An element with density 6 g cm-3 forms a fcc lattice with edge length of 4 x 10-8 cm. The molar mass of the element is (NA = 6 x 1023 mol-1) :(A) 57.6 g mol-1(B) 28.8 g mol-1(C) 82.6 g mol-1(D) 62 g mol-1

Answer»

Option : (A) 57.6 g mol^-1

We know that,

Number of atoms per unit cell of FCC = 4

Z = 4

We have given -

d = 6g/cm³,

a = edge length = 4 x 10^-8 cm

N_A = 6 x 10²³

d = \(\frac{ZM}{a^3N_A}\)

M = molar mass of element.

∴ 6g/cm³ = \(\frac{4\times M}{(4\times 10^{-8})^3cm^3\times 6\times 10^{23}mol^{-1}}\)

⇒ 6 x (64 x 10^-24) x 6 x 10²³ = 4M

⇒ M = \(\frac{6\times 6\times 64\times 10^{-1}}{4}g/mol\)

∴ M = 57.6 g/mol.

Discussion

27.	(I) How many unpaired electrons are present in the ground state of Fe3+ (z = 26), Mn2+ (z = 25) and argon (z = 18)? (II) Explain about the significance of de Broglie equation?
Answer» (I) Fe → Fe³⁺ + 3e^- Fe (Z = 26) Fe³⁺ = number of electrons = 23 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² for Fe atom. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ for Fe³⁺ ion. So, it contain 5 unpaired electrons. Mn (Z = 25) Electronic configuration is 1s² 2s² 2p⁵ 3s² 3p⁶ 4s² 3d⁵ Mn → Mn²⁺ + 2e^- Number of unpaired electrons in Mn²⁺ = 5 Ar (Z = 18). Electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ All orbitals are completely filled. So, no unpaired electrons in it. (II) Significance of de Broglie equation: 1. λ = \(\frac{h}{mv}\) This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle. 2. For a particle with high linear momentum (mv) the wavelength will be so small and cannot be observed. 3. For a microscopic particle such as an electron, the mass is of the order of 10^-31 kg, hence the wavelength is much larger than the size of atom and it becomes significant. 4. For the electron, the de Broglie wavelength is significant and measurable while for the iron ball it is too small to measure, hence it becomes insignificant.

27.

(I) How many unpaired electrons are present in the ground state of Fe3+ (z = 26), Mn2+ (z = 25) and argon (z = 18)? (II) Explain about the significance of de Broglie equation?

Answer»

(I) Fe → Fe³⁺ + 3e^-

Fe (Z = 26)

Fe³⁺ = number of electrons = 23

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² for Fe atom.

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ for Fe³⁺ ion.

So, it contain 5 unpaired electrons.

Mn (Z = 25) Electronic configuration is

1s² 2s² 2p⁵ 3s² 3p⁶ 4s² 3d⁵

Mn → Mn²⁺ + 2e^-

Number of unpaired electrons in Mn²⁺ = 5

Ar (Z = 18). Electronic configuration is

1s² 2s² 2p⁶ 3s² 3p⁶

All orbitals are completely filled. So, no unpaired electrons in it.

(II) Significance of de Broglie equation:

1. λ = \(\frac{h}{mv}\) This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle.

2. For a particle with high linear momentum (mv) the wavelength will be so small and cannot be observed.

3. For a microscopic particle such as an electron, the mass is of the order of 10^-31 kg, hence the wavelength is much larger than the size of atom and it becomes significant.

4. For the electron, the de Broglie wavelength is significant and measurable while for the iron ball it is too small to measure, hence it becomes insignificant.

Discussion

28.

(I) Mention any two anomalous properties of second period elements? (II) Arrange Na+ , Mg2+ and Al3+ in the increasing order of ionic radii. Give reason?

Answer»

(I) Anomalous properties of second period elements:

1. In the 1^st group, lithium forms compounds with more covalent character while the other elements of this group form only ionic compounds.

2. In the 2^nd group, beryllium forms compounds with more covalent character while the other elements of this family form only ionic compounds.

(II) Na⁺ , Mg²⁺ and Al³⁺ are isoelectronic cations.

Ions	Na⁺	Mg²⁺	Al³⁺
Number of electrons	10	10	10
Number of Nuclear charge	11	12	13

The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Hence the increasing order of ionic radii is, r_Na⁺, > r_Mg²⁺, >r_Al³⁺

Discussion

29.	In which of the following cases blood cells will shrink: (a) when placed in water containing more than 0.9% (mass/ volume) NaCl solution. (b) when placed in water containing less than 0.9% (mass /volume) NaCl solution. (c) when placed in water containing 0.9% (mass/volume) NaCl solution. (d) when placed in distilled water.
Answer» Correct answer is (a)When placed in water containing more than 0.9% (mass/ volume) NaCl solution because fluid inside blood cells is isotonic with 0.9% NaCl solution. When concentration of solute outside of the cell is increased then due to osmosis the solvent inside the cell come out thus blood cell get shrink. Answer is (a) when placed in water containing more than 0.9% (mass/ volume) NaCl solution. because fluid inside blood cells is isotonic with 0.9% NaCl solution

29.

In which of the following cases blood cells will shrink: (a) when placed in water containing more than 0.9% (mass/ volume) NaCl solution. (b) when placed in water containing less than 0.9% (mass /volume) NaCl solution. (c) when placed in water containing 0.9% (mass/volume) NaCl solution. (d) when placed in distilled water.

Answer»

Correct answer is (a)When placed in water containing more than 0.9% (mass/ volume) NaCl solution

because fluid inside blood cells is isotonic with 0.9% NaCl solution. When concentration of solute outside of the cell is increased then due to osmosis the solvent inside the cell come out thus blood cell get shrink.

Answer is (a) when placed in water containing more than 0.9% (mass/ volume) NaCl solution.

because fluid inside blood cells is isotonic with 0.9% NaCl solution

Discussion

30.	Which of the following is the weakest reducing agent in group 15?(A) NH3(B) PH3(C) A8H3(D) BiH3
Answer» Option : (A) NH₃ In group 15th element Hydrides have reducing character. The reducing character of Hydrides increases from Top to bottom. because the Z-H bond strength (Z = N,P, As, Bi) decreases down the group because of increase in the size of central atom. Hence, NH₃ is weak reducing agent.

30.

Which of the following is the weakest reducing agent in group 15?(A) NH3(B) PH3(C) A8H3(D) BiH3

Answer»

Option : (A) NH₃

In group 15th element Hydrides have reducing character. The reducing character of Hydrides increases from Top to bottom. because the Z-H bond strength (Z = N,P, As, Bi) decreases down the group because of increase in the size of central atom.

Hence,

NH₃ is weak reducing agent.

Discussion

31.	A certain reaction obeys the rate equation (in the integrated form) `[C^(1-n)-C_(@)^(1-n)]=(n-1) kt` where `[email protected]` is the initial concentration of C is the concentration after time, t Then:A. The unit of k for n=1 is `"sec"^(-1)`B. The unit of k for n=2 is litre `"mol"^(-1) "sec"^(-1)`C. The unit of k for n=3 is mol `"litre"^(-1) "sec"^(-1)`D. The unit of k for n=3 is `"lite"^2"mol"^2"sec"^(-1)`
Answer» Correct Answer - A,B,D Unit of `k=("mol"/"litre")^(1-n) xx s^(-1)` for n=1 is `sec^(-1)`, n=2 is litre `mol^(-1) sec^(-1)`, n=3 is `"litre"^2 mol^(-2) sec^(-1)`

Discussion

32.	Consider the following case of competing `1^(st)` order reactions. After the start of the reaction at t=0 with only A,the [C] is equal to the [D] at all time.The time in which all three concentrations wil be equal is given by :A. `t=1/(2k_1)ln3`B. `t=1/(2k_2)ln3`C. `t=1/(3k_1)ln2`D. `t=1/(3k_2)ln2`
Answer» Correct Answer - A,B `k_1=k_2` `implies 2/3rd` of A has reacted for [A] =[C]=[D] `:. K_1+k_2=1/t "In"([A]_0)/(1/3[A]_0)` `implies t=1/(k_1+k_2)"In" 3=1/(2k_1)"In"3=1/(2k_2)"In"3`

Discussion

33.	A certain reaction `A rarr B` follows the given concentration (Molarity)-time graph. Which of the following statements is/are true? A. The reaction is second order with respect to AB. The rate for this reaction at 20 second will be `7xx10^(-30) M s^(-1)`C. The rate for this reaction at 80 second will be `1.75xx10^(-3) M s^(-1)`D. The [B] will be 0.35 M at t=60 second
Answer» Correct Answer - B,D Rate for the reaction at 20 second `=0.35/50=7xx10^(-3) Ms^(-1)` `{:(,Ato,B),(t=0,0.4,0),(t=60,0.05,0.35):}`

Discussion

34.	100mL of 0.5 M `CuCO_4` (aq) solution was electrolysed using inert electrodes by passing current till the pH of the resulting solution was 2. The solution after electrolysis was Calculate the required volume (in mL) of `Na_2SO_2O_3`:A. `112.5ml`B. `100ml`C. `125ml`D. none of these
Answer» Correct Answer - A Initial `m`. Mole of `Cu^(2+)=5` `m`. Mole of `H^(+)` produced `=1` `m`. Mole of `Cu^(2+)` converted to `Cu=1/2=0.5` `m`. Mole of `Cu^(2+)` remaining in solution `=5-0.5=4.5` `2Cu^(2+)+4l^(-)toCu_(2)l_(2)+l_(2)` `l_(2)+2Na_(2)S_(2)O_(3)to2Nal+Na_(2)S_(4)O_(6)` `4.5=0.04xxV` `V=112.5` mol

Discussion

35.	Which among the following substances is used for bleaching ?(Oxygen, Hydrogen, Chlorine, Nitrogen)
Answer» Chlorine substances is used for bleaching.

Discussion

36.	The Molecular Formula of Glucose is -A. C2H5OH B. C6H12O6 C. H2SO4 D. C5H15N7
Answer» Option : B. C₆H₁₂O₆

Discussion

37.	Which is the Cation present in MgCl2 solution?
Answer» Magnesium ion/Mg²⁺

Discussion

38.	The aqueous solution of an acid conducts electricity because an acid is -A. soluble in water B. insoluble in water C. ionized in water D. none of these
Answer» Option : C. ionized in water

Discussion

39.	How many electrons are there in the outermost shell of lithium?A. 1 B. 2 C. 3 D. 4
Answer» Option : A. 1

Discussion

40.	A few facts regarding a gas is given(i) It is a poisonous gas(ii) It is formed by the incomplete burning of fuels.(a) Which is this gas? (b) What is the name of the mixture of this gas and Nitrogen?
Answer» (a) Carbon monoxide/CO (b) Producer gas

40.

A few facts regarding a gas is given(i) It is a poisonous gas(ii) It is formed by the incomplete burning of fuels.(a) Which is this gas? (b) What is the name of the mixture of this gas and Nitrogen?

Answer»

(a) Carbon monoxide/CO

(b) Producer gas

Discussion

41.	The element of zero group of the Periodic Table is –A. Cl B. H C. He D. Na
Answer» Option : C. He

Discussion

42.	Equation showing the reaction between Zinc and Hydrochloric acid is given Zn0 + 2H+Cl– → ZnXCl2 –1 + H20 (a) What is the oxidation state (X) of zinc in ZnCl2? (b) Which is the oxidizing agent in this reaction?
Answer» (a) +2 (b) HCl

Discussion

43.	Dime-Ten cents, Nickel-______?
Answer» Correct answer is Five cents

Discussion

44.	Which one of the following is the most electronegative element?A. F B. K C. I D. Na
Answer» Option : A. F

Discussion

45.	Which gas is deposited on Cathode in the electrolysis of water?A. Oxygen B. Ozone C. Hydrogen D. None of these
Answer» Option : C. Hydrogen

Discussion

46.	Analyse the symbol of the element given. (Symbol is not real) 31/15X(a) Write down its electronic configuration? (b) How many neutrons are present in this atom?
Answer» (a) Electronic configuration is 2, 8, 5 (b) Number of neutron = 31 – 15 = 16

46.

Analyse the symbol of the element given. (Symbol is not real) 31/15X(a) Write down its electronic configuration? (b) How many neutrons are present in this atom?

Answer»

(a) Electronic configuration is 2, 8, 5

(b) Number of neutron = 31 – 15 = 16

Discussion

47.	What is the pH value of our body?A. 2-3 B. 5-7 C. 7.0 – 7.8 D. 9.1 – 9.5
Answer» Option : C. 7.0 – 7.8

Discussion

48.	Choose the statements suitable to CO2 from those given below. (i) Used for making soft drinks. (ii) Used as a reducing agent in metallurgy (iii) Causes formation of Carboxy hemoglobin (iv) Used for making dry ice.
Answer» Statements (i) and (iv) are correct.

Discussion

49.	Which one of the following compounds is an acid?A. Na2O B. CuO C. Ca(OH)2 D. H2SO4
Answer» Option : D. H₂SO₄

Discussion

50.

Which property of sulphuric acid is exhibited in each of the reactions given below.(a) A few drops of Con. H2SO4 is added to sugar. (b) Cu is heated with Con. H2SO4 (c) Chlorine gas is collected in the laboratory by passing through Conc. H2SO4

Answer»

(a) Dehydrating nature

(b) Oxidizing property

(c) Non-volatile property

Discussion

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