Explore topic-wise InterviewSolutions in Current Affairs.

1. Phenylketonuria

2. Haemophilia

3. Down’s syndrome

4. Sickle-cell anaemia

(B) is wrong

Chemical or physical factors that induce mutation are referred as mutagens and increase in the whole set of chromosome is called poly ploidy.

It is given in the question that, 

Area of cross section of river = 45m × 2m = 90 m² 

Rate of flow = 3 km/hr

= \(\frac{3\times(1000\,m)}{(60\,m)}\)

= 50 m/min 

Therefore, 

Volume of water flowing through the cross-section in one minute = 90 m² × 50 m/min 

= 4500 m³ per minute

We know that if the unit place digit is divisible by 2 then our given number is divisible by 2.

Here, the unit place of the given number is 4 which is divisible by 2

∴The given number 94 is divisible by 2

Sickle cell anaemia is caused by the substitution of glutamic acid by valine, which is due to the single base substitution at the sixth codon of the beta-globin gene from GAG to GUG.

We know that, 

Volume of cuboid = Length × Breadth × Height 

Let the depth of the pit be x m 

Therefore, 

Volume = 5 × 3.5 × x 

It is given in the question that, 

Volume = 14 m³ 

Therefore, 

Depth, x = \(\frac{Volume}{Length\,\times\,Width}\)

x = \(\frac{14}{5\,\times\,3.5}\)

= 0.8 m 

= 80 cm

Given that, 

Volume of cylindrical tank = 1848 m³ 

Diameter = 14 m 

So, Radius = 7 m 

We know that, 

Volume of cylinder = \(\pi r^2h\)

1848 = \(\frac{22}{7}\times7\times7\times h\)

h = \(\frac{1848}{22\times7}\)

h = 12 m 

Hence, option B is correct

1. Down’s syndrome

2. 21^st trisomy/Non-disjunction of chromosomes during Anaphase I of meiosis.

Given that, 

Diameter of the circular well = 2 m 

Radius = 1 m 

Height = 14 m 

Therefore, 

Volume of cylindrical well = \(\pi r^2h\)

= \(\frac{22}{7}\times1\times1\times14\)

= 44 m³ 

Hence, option D is correct

(a) A-monosomy of sex chromosome B – Klinefelters syndrome 

(b) Failure of separation of homologous chromosomes during anaphase I of meosis lead to gain or loss of chromosome, it is called aneupoidy

It is given in the question that, 

Diameter = 7m 

Hence, 

Radius = 3.5 m 

Depth = 20 m 

Volume of the earth to be dug out = \(\pi r^2h\)

= \(\frac{22}{7}\times3.5\times3.5\times20\)

= 770 m³

Volume of the earth piled upon the given plot = 28× 11 × h 

Therefore,

Height = \(\frac{770}{28\times11}\)

= \(\frac{70}{28}\)

= 2.5 m

We know that if the unit place digit is divisible by 2 then our given number is divisible by 2.

Here, the unit place of the given number is 0 which is divisible by 2

∴The given number 570 is divisible by 2

We know that if the unit place digit is divisible by 2 then our given number is divisible by 2.

Here, the unit place of the given number is 5 which is not divisible by 2

∴The given number 285 is not divisible by 2

We know that if the unit place digit is divisible by 2 then our given number is divisible by 2.

Here, the unit place of the given number is 8 which is divisible by 2

∴The given number 2398 is divisible by 2

We know that if the unit place digit is divisible by 2 then our given number is divisible by 2.

Here, the unit place of the given number is 2 which is divisible by 2

∴The given number 79532 is divisible by 2

We know that if the unit place digit is divisible by 2 then our given number is divisible by 2.

Here, the unit place of the given number is 6 which is divisible by 2

∴The given number 13576 is divisible by 2

We know that if the unit place digit is divisible by 2 then our given number is divisible by 2.

Here, the unit place of the given number is 1 which is not divisible by 2

∴The given number 46821 is not divisible by 2

We know that if the unit place digit is divisible by 2 then our given number is divisible by 2.

Here, the unit place of the given number is 3 which is not divisible by 2

∴The given number 84663 is not divisible by 2

We know that if the unit place digit is divisible by 2 then our given number is divisible by 2.

Here, the unit place of the given number is 9 which is not divisible by 2

∴The given number 66669 is not divisible by 2

We know that if the unit place digit is divisible by 5 then our given number is divisible by 0 or 5.

Here, the unit place of the given number is 5 which is divisible by 5

∴The given number 95 is divisible by 5

We know that if the unit place digit is divisible by 5 then our given number is divisible by 0 or 5.

Here, the unit place of the given number is 0 which is divisible by 5

∴The given number 470 is divisible by 5

We know that if the unit place digit is divisible by 5 then our given number is divisible by 0 or 5.

Here, the unit place of the given number is 6 which is not divisible by 5

∴The given number 1056 is not divisible by 5

We know that if the unit place digit is divisible by 5 then our given number is divisible by 0 or 5.

Here, the unit place of the given number is 5 which is divisible by 5

∴The given number 2735 is divisible by 5

We know that if the unit place digit is divisible by 5 then our given number is divisible by 0 or 5.

Here, the unit place of the given number is 3 which is not divisible by 5

∴The given number 55053 is not divisible by 5

We know that if the unit place digit is divisible by 5 then our given number is divisible by 0 or 5.

Here, the unit place of the given number is 0 which is divisible by 5

∴The given number 35790 is divisible by 5

We know that if the unit place digit is divisible by 5 then our given number is divisible by 0 or 5.

Here, the unit place of the given number is 5 which is divisible by 5

∴The given number 98765 is divisible by 5

We know that if the unit place digit is divisible by 5 then our given number is divisible by 0 or 5.

Here, the unit place of the given number is 8 which is not divisible by 5

∴The given number 42658 is not divisible by 5

We know that if the unit place digit is divisible by 5 then our given number is divisible by 0 or 5.

Here, the unit place of the given number is 0 which is divisible by 5

∴The given number 77990 is divisible by 5

We know that if the unit place digit is divisible by 10 i.e. the unit place is having the value 0 then given number is divisible by 10.

Here, the unit place of the given number is 5 which is not divisible by 10

∴The given number 205 is not divisible by 10

We know that if the unit place digit is divisible by 10 i.e. the unit place is having the value 0 then given number is divisible by 10.

Here, the unit place of the given number is 0 which is divisible by 10

∴The given number 90 is divisible by 10

We know that if the unit place digit is divisible by 10 i.e. the unit place is having the value 0 then given number is divisible by 10.

Here, the unit place of the given number is 4 which is not divisible by 10

∴The given number 1174 is not divisible by 10

We know that if the unit place digit is divisible by 10 i.e. the unit place is having the value 0 then given number is divisible by 10.

Here, the unit place of the given number is 0 which is divisible by 10

∴The given number 57930 is divisible by 10

We know that if the unit place digit is divisible by 10 i.e. the unit place is having the value 0 then given number is divisible by 10.

Here, the unit place of the given number is 5 which is not divisible by 10

∴The given number 60005 is not divisible by 10

We know that the sum of digits of the number is divisible by 3 then the number is divisible by 3.

Sum of the digits is 11, which is not divisible by 3

∴The given number 83 is not divisible by 3

We know that the sum of digits of the number is divisible by 3 then the number is divisible by 3.

Sum of the digits is 18, which is divisible by 3

∴The given number 378 is divisible by 3

We know that the sum of digits of the number is divisible by 3 then the number is divisible by 3.

Sum of the digits is 15, which is divisible by 3

∴The given number 474 is divisible by 3

We know that the sum of digits of the number is divisible by 3 then the number is divisible by 3.

Sum of the digits is 19, which is not divisible by 3

∴The given number 1693 is not divisible by 3

Let a be the first term and d be the common difference. 

Then, T₈ = a + (8 – 1) d ⇒ 23 = a + 7d ...(i) 

T₁₀₂ = a + (102 – 1) d ⇒ 305 = a + 101d ...(ii) 

∴ Eqn (ii) – Eqn (i) 

⇒ 94d = 282 ⇒ d = 3 

Now substituting d = 3 in (i), we get 23 = a + 21 ⇒ a = 2. 

∴ a = 2, d = 3 

⇒ Series is 2 + 5 + 8 + 11 + .... .

We know that the sum of digits of the number is divisible by 3 then the number is divisible by 3.

Sum of the digits is 14, which is not divisible by 3

∴The given number 20345 is not divisible by 3

We know that the sum of digits of the number is divisible by 3 then the number is divisible by 3.

Sum of the digits is 21, which is divisible by 3

∴The given number 67035 is divisible by 3

We know that the sum of digits of the number is divisible by 3 then the number is divisible by 3.

Sum of the digits is 27, which is divisible by 3

∴ The given number 591282 is divisible by 3

We know that the sum of digits of the number is divisible by 3 then the number is divisible by 3.

Sum of the digits is 23, which is not divisible by 3

∴The given number 903164 is not divisible by 3

We know that the sum of digits of the number is divisible by 3 then the number is divisible by 3.

Sum of the digits is 3, which is divisible by 3

∴The given number 100002 is divisible by 3

Let the cost of 1 table be ‘a’ and cost of 1 chair be ‘b’.

Given,

4 tables and 3 chairs, together, cost Rs 2,250 and 3 tables and 4 chairs cost Rs 1950.

⇒ 4a + 3b = 2250 -------- (1)

Also,

3a + 4b = 1950 --------- (2)

Multiplying eq1 by 3 and eq2 by 4 and subtracting eq2 from eq1.

⇒ 12a + 9b – 12a – 16b = 6750 – 7800

⇒ 7b = 1050

⇒ b = 150

Substituting ‘b’ in eq1

⇒ 4a + 450 = 2250

⇒ a = Rs. 450

Cost of 2 chair and 1 table = 2b + a = Rs. 750

Let the cost of 1 bag be ‘a’ and 1 pen be ‘b’.

Given,

3 bags and 4 pens together cost Rs 257 whereas 4 bags and 3 pens together cost Rs 324.

⇒ 3a + 4b = 257 ----------- (1)

and 4a + 3b = 324 ---------- (2)

Multiplying eq1 by 4 and eq2 by 3 and subtracting eq2 from eq1.

⇒ 12a + 16b – 12a – 9b = 1028 – 972

⇒ 7b = 56

⇒ b = Rs. 8

Substituting in eq1

⇒ 3a + 32 = 257

⇒ a = Rs. 75

Cost of 1 bag and 10 pens = 75 + 10 × 8 = Rs. 155

Let’s assume the cost of 1 table is ₹ x and cost of 1 chair is ₹ y. 

Then, according to the question 

4x + 3y = 2250 … (i) 

3x + 4y = 1950 … (ii) 

On multiplying (i) with 3 and (ii) with 4, 

We get, 

12x + 9y = 6750 … (iii) 

12x + 16y = 7800 … (iv) 

Now, subtracting equation (iv) from (iii), We get, 

-7y = -1050 

y = 150 

Using y = 150 in (i), we find x 

4x + 3(150) = 2250 

4x = 2250 – 450 

x = 1800/ 4 

⇒ x = 450 

From the question, it’s required to find the value of (x + 2y) 

⇒ 450 + 2(150) = 750 

Therefore, the total cost of 2 chairs and 1 table is ₹ 750.

Let the cost of a bag and a pen be ₹ x and ₹ y, respectively. 

Then, according to the question 

3x + 4y = 257 … (i) 

4x + 3y = 324 … (ii) 

On multiplying equation (i) by 3 and (ii) by 4, 

We get, 9x + 12y = 770 … (iii) 

16x + 12y = 1296 … (iv) 

Subtracting equation (iii) from (iv), we get 

16x – 9x = 1296 – 771 

7x = 525 

x = 525/7 = 75 

Hence, the cost of a bag = ₹ 75 

Substituting x = 75 in equation (i), We get, 

3 x 75 + 4y = 257 

225 + 4y = 257 

4y = 257 – 225 

4y = 32 

y = 32/4 = 8 

Hence, the cost of a pen = ₹ 8 

From the question, it’s required to find the value of (x + 10y) 

⇒ 75 +10(8) = 20 

Therefore, the total cost of 1 bag and 10 pens = 75 + 80 = ₹ 155.

We know that the sum of digits of the number is divisible by 9 then the number is divisible by 9.

Sum of the digits is 12, which is not divisible by 9

∴The given number 327 is not divisible by 9

We know that the sum of digits of the number is divisible by 9 then the number is divisible by 9.

Sum of the digits is 18, which is divisible by 9

∴The given number 7524 is divisible by 9

We know that the sum of digits of the number is divisible by 9 then the number is divisible by 9.

Sum of the digits is 9, which is divisible by 9

∴The given number 32022 is divisible by 9