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(i)

(ii) 

Let the duration of each period, when there are 9 periods a day in the school, be x minutes. The following table is obtained.

If there is more number of periods a day in the school, then the duration of each period will be lesser. Hence, this is a case of inverse proportion. 

Therefore 45 × 8 = x × 9

x = (45 x 8)/9 = 40

Hence, in this case, the duration of each period will be 40 minutes.

Let the actual length be x cm. When the bacteria is enlarged this much its length becomes 5 cm.

Then, \(x\times50000=5.\)

\(\therefore x=\frac{1}{1000}=\frac{1}{10^4}=10^{-4}\).

(i) These are in inverse proportion because if there are more workers, then it will take lesser time to complete that job.

(ii) No, these are not in inverse proportion because in more time, we may cover more distance with a uniform speed.

(iii) No, these are not in inverse proportion because in more area, more quantity of crop may be harvested.

(iv) These are in inverse proportion because with more speed, we may complete a certain distance in a lesser time.

(v) These are in inverse proportion because if the population is increasing, then the area of the land per person will be decreasing accordingly.

(i) The number of workers and the time to complete the job is in inverse proportion because less workers will take more time to complete a work and more workers will take less time to complete the same work. 

(ii) Time and distance covered in direct proportion. 

(iii) It is a direct proportion because more are of cultivated land will yield more crops. 

(iv) Time and speed are inverse proportion because if time is less, speed is more. 

(v) It is a inverse proportion. If the population of a country increases, the area of land per person decreases.

(i) Let the number of days required by 1 man to fit all the windows be x. The following table is obtained.

Lesser the number of persons, more will be the number of days required to fit all the windows. Hence, this is a case of inverse proportion. Therefore, 2 × 3 = 1× x

x = 6

Hence, the number of days taken by 1 man to fit all the windows is 6.

(ii) Let the number of persons required to fit all the windows in one day be y. The following table is formed.

Lesser the number of days, more will be the number of persons required to fit all the windows. Hence, this is a case of inverse proportion. Therefore, 2 × 3 = y × 1 

y = 6 

Hence, 6 persons are required to fit all the windows in one day.

1. Number of chairs and the number of spectators. 

2. Quantity (litres) of water and number of vessels required to store the water.

Total investment = Rs 360000 

Profit earned = 12% 

∴ Total profit = 12% of 360000 

= 12/100 x 360000 

= Rs 43200 

Proportion of investment = 3 : 4 : 7 : 6

Let the investment of Jaya, Seema, 

Nikhil and Neelesh be Rs 3x, Rs 4x, Rs 7x and Rs 6x respectively. 

3x + 4x + 7x + 6x = 360000 

∴ 20x = 360000 

∴ x = 360000/20

= 18000 

∴ Jaya’s investment = 3x = 3 x 18000 = Rs 54000 

Also, profit made by them is Rs 43200 

∴ 3x + 4x + 7x + 6x = 43200 

∴ 20x = 43200 

∴ x = 43200/20

= 2160 

∴ Nikhil’s profit = 7x = 7 x 2160 = Rs 15120 

∴ Jaya’s share in the capital was Rs 54000 and the profit made by Nikhil was Rs 15120.

Yes. 

If amount of petrol filled in the motorcycle is less, it will travel less distance and if the amount of petrol filled is more, it will travel more distance.

Hence, the amount of petrol filled in the motorcycle and the distance traveled by it are in direct proportion.

Let the food supplement required for 12 cows be x kg. 

The quantity of food supplement required and the number of cows are in direct proportion. 

∴ (13kg500gram)/9 = xkg./12

∴ 13.5/9 = x/12 ….(13 kg 500 gram = 13.5 kg) 

∴ 13.5 × 12 = 9x ….(Multiplying both sides by 9 x 12) 

∴ (13.5 x 12)/9 = x 

∴ x = 18 

∴ The food supplement required for 12 cows is 18 kg.

Total investment = Rs 80000 

Nachiket’s investment = Total investment – (Shweta’s investment + Piyush’s investment) 

= 80000 – (30000+ 12000) 

= 80000 – 42000 = Rs 38000

Profit earned = 24% 

∴ Total profit = 24% of 80000 

= 24/100 x 80000 

= Rs 19200 

Proportion of investment = 30000 : 12000 : 38000 

= 15 : 6 : 19 …. (Dividing by 2000) 

Let the profit of Shweta, Piyush and Nachiket be Rs 15x, Rs 6x and Rs 19x respectively. 

15x + 6x + 19x = 19200 

∴ 40x = 19200 

∴ x = 19200/40 = 480

∴ Nachiket’s profit = 19x = 19 × 480 = Rs 9120 

∴ Nachiket’s investment is Rs 38000 and his profit is Rs 9120.

Total investment = Rs 2400 + Rs 5200 + Rs 3400 = Rs 11000 

Total profit = 50% of 11000 

= 50/100 × 11000 = Rs 5500

Proportion of shares = 2400 : 5200 : 3400 

= 12 : 26 : 17 …. (Dividingby 200) 

Let the profit of Saritaben, Ayesha and Meenakshi be Rs 12x, Rs 26x and Rs 17x respectively. 

12x + 26x + 17x = 5500 

∴ 55x = 5500 

∴ x = 100

∴ Saritaben’s profit = 12x = 12 × 100 = Rs 1200 

Ayesha’s profit = 26x = 26 × 100 = Rs 2600 

Meenakshi’s profit = 17x = 17 × 100 = Rs 1700 

∴ Saritaben’s new investment = 2400 + 1200 = Rs 3600 

Ayesha’s new investment = 5200 + 2600 = Rs 7800 

Meenakshi’s new investment = 3400 + 1700 = Rs 5100 

∴ New proportion of shares = 3600 : 7800 : 5100 

= 12 : 26 : 17 …. (Dividing by 300) 

∴ The proportion of the shares in the capital during the following year is 12 : 26 :17

Proportion of share = 3:7 

Let the profits of A and B be Rs 3x and Rs 7x respectively. 

3x + 7x = 24500 

∴ 10x = 24500 

∴ x = 24500/10 = 2450

∴  Profit earned by A = 3x = 3 × 2450 = Rs 7350 

Amount given by A = 2% of his profit 

= 2/100 × 7350 = Rs 147 

Profit earned by B = 7x = 7 × 2450 = Rs 17150 

Amount given by B = 2% of his profit

= 2/100 × 17150 

= Rs 343 

∴ The amount given by A and B to the Soldiers’ Welfare Fund are Rs 147 and Rs 343 respectively.

Total investment = Rs 144000 

Profit earned = 20% 

∴ Total profit = 20% of 144000 

= 20/100 x 144000

= Rs 28800 

Proportion of investment of Suresh and Ramesh = 4:5

Let the profit of Suresh be Rs 4x and that of Ramesh be Rs 5x.

4x + 5x = 28800 

∴ 9x = 28800 

∴ x = 28800/9 

= 3200 

∴ Suresh’s profit = 4x = 4 × 3200 = Rs 12800 

Ramesh’s profit = 5x = 5 × 3200 = Rs 16000 

∴ The profit earned by Suresh and Ramesh are Rs 12800 and Rs 16000 respectively.

Diameter of the table cover = 1.5m

Circumference of the table cover 

2πr – πd = 3.14 × 1.5 (∵ d = 2r) = 4.710 mts

∴ Length of the lace required = 4.71 mt

Cost of lace per meter = Rs. 15

∴ Total cost of the lace =4.71 × 15

= Rs 70.65

According to the question,

Area of the circular playground = 22176 m² (Given)

Let r be the radius of the circle.

∴ πr² = 22176

⇒ (22/7)r² = 22176

⇒ r² = 22176 × (22/7)

⇒ r² = 7056

⇒ r = 84

∴ Radius of the circular playground = 84 m

Now, circumference of the circle = 2πr

= 2×(22/7)×84

= 528 m

Cost of fencing 1 meter of ground = Rs 50

∴ Cost of fencing the total ground = Rs 528 × 50 = Rs 26,400

Given,

Area of the circular playground = 22176 m²

And the cost of fencing per metre = ₹50

If the radius of the ground is taken as r.

Then, its area = πr²

πr² = 22176

r² = 22176(7/22) = 7056

Taking square root on both sides, we have

r = 84 m

We know that, fencing is done only on the circumference of the ground

Circumference of the ground = 2πr = 2(22/7)84 = 528 m

So, the cost of fencing 528 m = ₹50 x 528 = ₹26400

Therefore, the cost of fencing the ground is ₹26400.

The correct answer is (i) 11/70  (ii) 1/10

The correct answer is 7/8,6/7,4/5,3/4

The correct answer is 7/8,6/7,4/5,3/4

The correct answer is 6 3/100

The value of 50 coins of 50 paisa = Rs 25

The correct answer is 3/5

Correct option is  D) 80.815

Correct option is : (c) 50

2, 5, 10, 17, 26, 37, ………. 

(1 × 1 + 1), (2 × 2 + 1), (3 × 3 + 1), (4 × 4 + 1), 

(5 × 5 + 1), (6 × 6 + 1), …………. 

So, next number is 

(7 × 7 + 1) = 50

8.76 – 2.68 = 6.08

The correct answer is (D)

Correct option is (d) 106

1, 6, 16, 31, 51, 76, ………….. 

1, (1 + 5), (6 + 10), (16 + 15), (31 + 20), 

(51 + 25), (76 + 30), …………… 

∴ So, next number is (76 + 30) = 106.

The mode of some observations is ‘x’. If ‘3’ is subtracted from every observation then the mode = x – 3

The given observations are 10, 12, 11, 10, 15, 20, 19,21,11,9, 10

∴ The most frequently occurring observation is 10. 

∴ Mode = 10

Yes, we can identify the frequency of observation with a frequency polygon at a particular point. Since the height of the polygon is equal to frequency of polygon.

If one or two or more observations equal to mode are included there will be no change in the mode. 

Ex: The mode of 5, 6, 7, 8, 7, 9 is 7. If 3, 7’s are added to above observations there will be no change in the mode.

Correct option is  A) 5

Correct option is  B) 3

Distance covered for 1 lit, of petrol = 62.5 km 

∴ Distance covered for 10 lit, of petrol = 62.5 × 10 = 625 km

Correct option is  B) \(\frac{n+1}{2}\)

Correct option is  D) No mode

Correct option is  A) 4.6

Correct option is  D) None

Correct option is  A) Mean

 correct answer is (A) R.

The correct answer is (B) I 

Correct option is  D) None

Correct option is  D) 3, 7

Correct option is  D) 24

Correct option is  B) 2

Correct option is  B) 6x

(B) The new alphabet series is :

M L K J I H G F E D C B N A P Q R S T U V W X Y Z

Nineteenth letter from the right will be letter F.