## NIELIT Scientist -B CS (22-07-17)

Question 1 |

What is the maximum number of the handshakes that can happen in the room with 5 people in it?

15 | |

10 | |

6 | |

5 |

Question 1 Explanation:

Let’s suppose there is ‘n’ number of persons are there and each one suppose to do an handshake with ‘n-1 persons.

So, Total number of handshakes is = n(n-1) (with repetition)

But we need without repetition means distant handshakes (A→ B, B → A) so we divided by ‘2’

= n(n-1) / 2

= 5(5-1) / 2

= 10

So, Total number of handshakes is = n(n-1) (with repetition)

But we need without repetition means distant handshakes (A→ B, B → A) so we divided by ‘2’

= n(n-1) / 2

= 5(5-1) / 2

= 10

Question 2 |

A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?

120 meters | |

180 meters | |

324 meters | |

150 meters |

Question 2 Explanation:

Let’s take length of the train = ‘x’

Speed of the Train = 60km/hr and crosses a pole in 9 Seconds.

Speed of the train is km/hr and crosses a pole in seconds.

So we need to convert the km/hr into mts /sec.

For that one we need to multiply the speed of the train to (5/18).

Speed of the train = 60 * (5/18)

We have one Basic formula in Time and Distance

Distance = Speed * Time

x = 60 * (5/18) * 9

x = 150 mts

Speed of the Train = 60km/hr and crosses a pole in 9 Seconds.

Speed of the train is km/hr and crosses a pole in seconds.

So we need to convert the km/hr into mts /sec.

For that one we need to multiply the speed of the train to (5/18).

Speed of the train = 60 * (5/18)

We have one Basic formula in Time and Distance

Distance = Speed * Time

x = 60 * (5/18) * 9

x = 150 mts

Question 3 |

The percentage profit earned by selling an article for Rs. 1,920 is equal to the percentage loss incurred by selling the same article for Rs. 1,280. At what price should the article be sold to make 25% profit?

Rs 2,000 | |

Rs 2,200 | |

Rs 2,400 | |

Data inadequate |

Question 3 Explanation:

For this Problem, We need two important formulas.

1. %Profit = ((Selling Price - Cost Price) / (Cost Price) ) * 100

2. %Loss = ((Cost Price - Selling Price) / (Cost Price)) * 100

From the above question we know

% Profit = % Loss

Selling price when they got profit = 1920 /-

Selling price when they got loss = 1280 /-

((1920 - Cost Price ) / ( Cost Price) ) * 100 = (( Cost Price - 1280 ) / (Cost Price)) * 100

After Solving this

2 * Cost Price = 3200

Cost Price = 1600

We got cost price, Now we need to find out Selling price when they got 25% Profit

Cost Price = 1600

Profit% = 25

Selling Price = ?

%Profit = ((Selling Price - Cost Price) / (Cost Price) ) * 100

25 = ((Selling Price - 1600) / 1600)) * 100

25 * 16 = Selling Price - 1600

Selling Price = 1600 + 400

Selling Price = 2000/-

1. %Profit = ((Selling Price - Cost Price) / (Cost Price) ) * 100

2. %Loss = ((Cost Price - Selling Price) / (Cost Price)) * 100

From the above question we know

% Profit = % Loss

Selling price when they got profit = 1920 /-

Selling price when they got loss = 1280 /-

((1920 - Cost Price ) / ( Cost Price) ) * 100 = (( Cost Price - 1280 ) / (Cost Price)) * 100

After Solving this

2 * Cost Price = 3200

Cost Price = 1600

We got cost price, Now we need to find out Selling price when they got 25% Profit

Cost Price = 1600

Profit% = 25

Selling Price = ?

%Profit = ((Selling Price - Cost Price) / (Cost Price) ) * 100

25 = ((Selling Price - 1600) / 1600)) * 100

25 * 16 = Selling Price - 1600

Selling Price = 1600 + 400

Selling Price = 2000/-

Question 4 |

A boatman goes 2 Km against the current of the stream in 1 hour and goes 1 Km along the current in 10 minutes. How long will it take to go 5 Km in stationary water?

40 minutes | |

1 hour | |

1 hr 15 min | |

1 hr 30 min |

Question 4 Explanation:

Speed of the Down Stream = 2 km/ 1 = 2 km/hr

Speed of the Upstream = 5km / (10/60) = 6km/hr

Speed of the Boat still in water = (Downstream + Upstream ) / 2 = (6 + 2) / 2

= 4 km/hr

Time taken for 5 km in Stationary water = 5/4 = 1(1⁄4)

= 1Hr 15 Min

Speed of the Upstream = 5km / (10/60) = 6km/hr

Speed of the Boat still in water = (Downstream + Upstream ) / 2 = (6 + 2) / 2

= 4 km/hr

Time taken for 5 km in Stationary water = 5/4 = 1(1⁄4)

= 1Hr 15 Min

Question 5 |

The present ages of three persons in proportions 4 : 7 : 9. Eighty years ago, the sum of their ages was 56. Find their present ages (in years).

8, 20, 28 | |

16, 28, 36 | |

20, 35, 45 | |

None of the above |

Question 5 Explanation:

Present age of 3 persons is = 4x, 7x, 9x

Eight years ago their sum is = 56

So,

4x+7x+9x-(8+8+8) = 56 (Eight Years Ago so that we subtract their individual ages from present age)

20x = 80

x = 4

Present ages = 4 * 4 , 7*4, 9*4

= 16, 28, 36

Eight years ago their sum is = 56

So,

4x+7x+9x-(8+8+8) = 56 (Eight Years Ago so that we subtract their individual ages from present age)

20x = 80

x = 4

Present ages = 4 * 4 , 7*4, 9*4

= 16, 28, 36

Question 6 |

A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is?

6 hours | |

10 hours | |

15 hours | |

30 hours |

Question 6 Explanation:

Let’s take the first pipe to a fill a tank in ‘x’ hours.

So, from that question we know that second pipe fills a tank 5 hours faster than 1st pipe means second pipe fills a tank in ‘x/5’ hours.

And Second pipe fills a tank 4 hours slower than third pipe means third pipe can fill a tank in ‘x/20’ hours.

The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone.

(1 / x) + (5 / x) = (20 / x)

x = 15 Hours

So, from that question we know that second pipe fills a tank 5 hours faster than 1st pipe means second pipe fills a tank in ‘x/5’ hours.

And Second pipe fills a tank 4 hours slower than third pipe means third pipe can fill a tank in ‘x/20’ hours.

The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone.

(1 / x) + (5 / x) = (20 / x)

x = 15 Hours

Question 7 |

A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq feet, how many feet of fencing will be required?

34 | |

40 | |

68 | |

88 |

Question 7 Explanation:

Given that length and area, so we can find the breadth.

Length x Breadth = Area

20 x Breadth = 680

Breadth = 34 feet

Area to be fenced = 2B + L = 2 (34) + 20 = 88 feet

Length x Breadth = Area

20 x Breadth = 680

Breadth = 34 feet

Area to be fenced = 2B + L = 2 (34) + 20 = 88 feet

Question 8 |

In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?

810 | |

1440 | |

2880 | |

50400 |

Question 8 Explanation:

In ‘CORPORATION’ We have total 5 vowels (OOAIO) and taken has one unit

Remaining Consonants are (CRPRTN) 6 and add vowels unit 1.

6 + 1 = 7 In this R is repeated 2 times

In ‘CORPORATION’ We have total 5 vowels (OOAIO) and taken has one unit Remaining Consonants are (CRPRTN) 6 and add vowels unit 1. 6 + 1 = 7 In this R is repeated 2 times So number of ways we can arrange these words = 7! / 2! = 2520

number of ways we can arrange vowels is 5!

2 ‘O’ are repeated so 5! / 2! = 20

Total Number ways = 2520 * 20 = 50400

So number of ways we can arrange these words = 7! / 2! = 2520

number of ways we can arrange vowels is 5!

2 ‘O’ are repeated so 5! / 2! = 20

Total Number ways = 2520 * 20 = 50400

Remaining Consonants are (CRPRTN) 6 and add vowels unit 1.

6 + 1 = 7 In this R is repeated 2 times

In ‘CORPORATION’ We have total 5 vowels (OOAIO) and taken has one unit Remaining Consonants are (CRPRTN) 6 and add vowels unit 1. 6 + 1 = 7 In this R is repeated 2 times So number of ways we can arrange these words = 7! / 2! = 2520

number of ways we can arrange vowels is 5!

2 ‘O’ are repeated so 5! / 2! = 20

Total Number ways = 2520 * 20 = 50400

So number of ways we can arrange these words = 7! / 2! = 2520

number of ways we can arrange vowels is 5!

2 ‘O’ are repeated so 5! / 2! = 20

Total Number ways = 2520 * 20 = 50400

Question 9 |

Seats for Mathematics, Physics and Biology in a school are in the ratio of 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?

2 : 3 : 4 | |

6 : 7 : 8 | |

6 : 8 : 9 | |

None of the above |

Question 9 Explanation:

Seats for Mathematics, Physics and Biology in a school are in the ratio is 5:7:8.

There is a proposal to increase these seats by 40%, 50% and 75% respectively means

After increasing seats ratio will be

5 * (140/100) : 7 * (150/100) : 8 * (175/100)

5 * 140 : 7 * 150 : 8 * 175

28 : 42 : 56

2 : 3 : 4

There is a proposal to increase these seats by 40%, 50% and 75% respectively means

After increasing seats ratio will be

5 * (140/100) : 7 * (150/100) : 8 * (175/100)

5 * 140 : 7 * 150 : 8 * 175

28 : 42 : 56

2 : 3 : 4

Question 10 |

A flag staff 17.5 m high casts a shadow of length 40.25 m. The height of the building, which casts a shadow of length 28.75 m under similar condition will be?

10 m | |

12.5 m | |

17.5 m | |

21.25 m |

Question 10 Explanation:

Let’s take height of the building is x.

From that question we know that less height less shadow so it’s directly proportional.

40.25 : 17.5 : : 28.75 : x

x = (17.5 * 28.75) / 40.25

x = 12.5 m

From that question we know that less height less shadow so it’s directly proportional.

40.25 : 17.5 : : 28.75 : x

x = (17.5 * 28.75) / 40.25

x = 12.5 m

Question 11 |

Find the odd one in the following series for

396, 462, 572, 427, 671, 264

396, 462, 572, 427, 671, 264

396 | |

427 | |

671 | |

264 |

Question 11 Explanation:

Except 427 remaining numbers are sum of first number and last number is middle number.

Question 12 |

Find the odd one in the following series for

2, 5, 10, 17, 26, 37, 50, 64

2, 5, 10, 17, 26, 37, 50, 64

50 | |

26 | |

37 | |

64 |

Question 12 Explanation:

From the above sequence we know that

2 + 3 = 5

5 + 5 = 10

10 + 7 =17

17 + 9 = 26

26 + 11 = 37

37 + 13 = 50

50 + 15 = 65 But in the question they given 64. So 64 is wrong.

2 + 3 = 5

5 + 5 = 10

10 + 7 =17

17 + 9 = 26

26 + 11 = 37

37 + 13 = 50

50 + 15 = 65 But in the question they given 64. So 64 is wrong.

Question 13 |

The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs.) is

625 | |

630 | |

640 | |

650 |

Question 13 Explanation:

The difference between compound interest and simple interest on Rs. P for 2 years at R% per annum is = p * (R/100)^2

P * (R/100)^2 = 1

P * (4/100)^2 = 1

P * (1/25)^2 = 1

P = 625

So, Sum is 625

P * (R/100)^2 = 1

P * (4/100)^2 = 1

P * (1/25)^2 = 1

P = 625

So, Sum is 625

Question 14 |

The angle of elevation of the sun, when the length of the shadow of a tree is 3 times the height of the tree, is

30 ^{0} | |

45 ^{0} | |

60 ^{0} | |

90 ^{0} | |

None of the above |

Question 15 |

Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is

4 | |

5 | |

6 | |

8 |

Question 15 Explanation:

For this type of questions we have to find gcd of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

So, = H.C.F. of 3360, 2240 and 5600 = 1120.

sum = 1+1+2+0 = 4.

So, = H.C.F. of 3360, 2240 and 5600 = 1120.

sum = 1+1+2+0 = 4.