RRB NTPC Mathematics Test - 3 - RRB NTPC/ASM/CA/TA MCQ

Population growth calculation

Given:

Initial population = 66000

Increase by 5% in the first year

Decrease by 5% in the second year

Concept:
Percentage change in population

⇒ Population after first year = 66000 * 1.05

⇒ Population after second year = 66000 * 1.05 * 0.95

⇒ Final population = 65835

Therefore, Final population of the town is 65835.

Given:
The ratio of milk and water in mixture = 9 : 5
The new ratio after 6 liters of water added = 9 : 7
Calculation:
Let the quantity of milk and water in mixture initially was 9X and 5X respectively.
According to the question

⇒ 63X = 45X + 54
⇒ 63X - 45X = 54
⇒ 18X = 54
⇒ X = 3
The quantity of water initially = 3 × 9 = 27 liters
∴ The quantity of milk in the mixture initially was 27 liters.

Given:

The ratio of milk and water in the first vessel = 8 : 7

The ratio of milk and water in the second vessel = 7 : 9

The ratio of water and milk in the resultant mixture = 9 : 8

Calculation:

Let x litre of the first mixture and y litre of the second mixture are mixed.

Quantity of milk in x litre of the first mixture = 8x/15

Quantity of milk in y litre of the second mixture = 7y/16

Total quantity of the resultant mixture = (x + y)

Quantity of milk in (x + y) litre of the resultant mixture = 8(x + y)/17

8x/15 + 7y/16 = 8(x + y)/17

⇒ 8x/15 + 7y/16 = 8x/17 + 8y/17

⇒ 8x/15 – 8x/17 = 8y/17 – 7y/16

⇒ (136x – 120x)/15 × 17 = (128y – 119y)/17 × 16

⇒ 16x/15 = 9y/16

⇒ 256x = 135y

⇒ x/y = 135/256

∴ The required ratio is 135 : 256

Alternative Method:

The concentration of milk in the first mixture = 8/15

The concentration of milk in the second mixture = 7/16

The concentration of milk in the resultant mixture = 8/17

By the rule of Allegation,

⇒ 9/272 : 16/255

⇒ 9 × 255 : 16 × 272

⇒ 9 × 15 : 16 × 16

⇒ 135 : 256

∴ The required ratio is 135 : 256.

Given:

The third proportion to 10 and x is 90 and the third proportion of 12 and y is 27

Concept used:

If a, b and c are in proportion,

Then we can say that, b² = ac

Calculartion:

As per the question,

x² = 90 × 10

⇒ x = 30

Again,

As per the question,

y² = 27 × 12

⇒ y = 18

So value of (x + y) = 30 + 18

⇒ 48

∴ The correct option is C

Given:

SI = 2/25 × Principal

Time (T) = 2 × Rate (R)

Formula Used:

SI = (P × R × T) ÷ 100

P = Principal, R = Rate, T = Time

Calculation:

Let the Principal be 50 rupees

⇒ SI = 2/25 × 50 = 4 rupees

According to the question

⇒ R² = 4

⇒ R = 2%

∴ The rate of interest is 2% per annum.

Formula used:

N^th term of a series = a + (n – 1) × d

Where a is the first term of the series and d is the common difference

Calculation:

19^th term will be 94

Here 19^th term = 22 + (19 – 1) × 4

⇒ 94

∴ Required 19^th term = 94

Given

The total number of children in 2016 = 7500

The total number of children in 2017 = 10500

The Percentage (%) of children who are not overweight in 2016 = (100 - 30) % = 70%

The Percentage (%) of children who are not overweight in 2017 = (100 - 28) % = 72%

The number of children who were not overweight in the year 2016:

⇒ 70% of 7500 = 5250

The number of children who were not overweight in the year 2017:

⇒ 72% of 10500 = 7560

The total number of children who were not overweight in the year 2016 and 2017 together:

⇒ 5250 + 7560 = 12810

Given:

The shadow of a pole 6 m high and 15 m long.

At the same time, the shadow of a tree is 25 m long.

Calculation:

Let the height of the tree be H.

According to the question

⇒ 6/H = 15/25

⇒ 15H = 150

⇒ H = 10 m

∴ The required height of the tree is 10 m.

Given:

P(E) = 0.3

P(F) = 0.2

P(F/E) = 0.6

Calculation:

P(F/E) = [P(E∩F)/P(E)]

⇒ 0.6 = [P(E∩F)/0.3]

⇒ P(E∩F) = 0.18

Now,

P(E/F) = [P(E∩F)/P(F)]

⇒ P(E/F) = 0.18/0.2

⇒ P(E/F) = 0.9

∴ The required answer is 0.9.

Given:

The sum of radii of two spheres = 28 cm

and the ratio of volumes = 27:64

Formula used:

volume of sphere = 4/3 πr³

Calculation:

Let the radii of spheres be R₁ and R₂

and the volumes be V₁ and V₂

∴ R₁ + R₂ = 28

& V₁ / V₂ = 27/64

⇒ 4/3 π R₁³ / 4/3 π R₂³ = 27/64

⇒ (R₁ / R₂)³ = 27/64 = (3/4)³

⇒ R₁ / R₂ = 3/4

or R₁ : R₂ = 3:4

let the ratio be x

then, 3x + 4x = 28

or 7x = 28 ⇒ x = 4

∴ R₁ = 3x = 12 cm

R₂ = 4x = 16 cm

∴The radii are 12 cm and 16 cm

Given data:

Seat numbers in each class,

Occupancy percentage in each class,

Price per seat in each class.

Concept used:

Total income = (No. of seats × Occupancy rate × Price per seat)

⇒ First class income = 50 seats × 30/100 × 2000 Rs = 30,000 Rs

⇒ Second class income = 100 seats × 50/100 × 1000 Rs = 50,000 Rs

⇒ Third class income = 150 seats × 100/100 × 500 Rs = 75,000 Rs

⇒ Income ratio per category = 30,000 ∶ 50,000 ∶ 75,000 = 6 ∶ 10 ∶ 15

∴ Option A is the correct answer.

Given:

a = 2

b = 3

Concept:

We need to find the third proportional to a³ + b³ and a² + ab + b².

⇒ Substitute a and b into the two expressions to get the first and second numbers.

⇒ First number = a³ + b³ = 2³ + 3³ = 8 + 27 = 35

⇒ Second number = a² + ab + b² = 2² + 2 * 3 + 3² = 4 + 6 + 9 = 19

⇒ The third proportional (T) to two numbers (x and y) is given by the formula T = (y²)/x

So, substituting the first and second numbers:

⇒ T = (19²)/35 = 10.31

Therefore, the third proportional to a³ + b³ and a² + ab + b², when a = 2 and b = 3, is approximately 10.31 (correct to two decimal places).

Given:

Ram and Lal have a combined salary of Rs. 80,000

Concept:

Salary = Expenditure + Savings

Calculation:

Ram saves 10% of his salary

Lal saves 30% of his salary

It is given that above savings are equal

10% of x = 30% of (80000 – x)

[Here x is the salary of Ram thus salary of Lal will be (80000 – x)]

⇒

⇒

⇒

Ram’s salary is Rs. 60000

Lal’s salary is Rs. 20000

Given:

Container P is containing 50 L of milk

Container Q is containing 32 L of water.

From container P, 10 Litre of milk is taken out and poured into container Q.

Then, 8 Litre of mixture (milk and water) is taken out from container Q and poured into container P.

Calculation:

Initial quantity of milk in container P = 50 L

Initial quantity of water in container Q = 32 L

Now, 10 L of milk is taken out from container P,

So, Quantity of milk is remain in container P = 50 - 10

⇒ 40 L

Quantity of milk in container Q = 10 L

Quantity of water in container Q = 32 L

(Water : Milk) in container Q = 32 : 10

⇒ 16 : 5

⇒ Total mixture = 21 unit

Now, 8 Litre of mixture (milk and water) is taken out from container Q and poured into container P.

Quantity of milk in 8 L = 8 × 5/21

⇒ 40/21

Quantity of water in 8 L = 8 × 16/21

⇒ 128/21

Now, Quantity of milk in container P = 40 + 40/21

⇒ 880/21

Quantity of water in container Q = 32 - 128/21

⇒ 544/21

Required ratio = 880/21 : 544/21

⇒ 55 : 34

∴ Required ratio is 55 : 34.

Given, table worth Rs. 300 is offered with 20% and 10% off.

1^st Discount: 20% of Rs.300 = 60

After 1^st Discount

Selling price = 300 – 60 = Rs. 240

2^nd Discount: 10% of Rs.240 = 24

After 2^nd Discount

Selling price = 240 – 24 = Rs.216

Now an additional discount of 5% is given.

3^rd Discount: 5% of Rs.216 = 10.8

After 3^rd Discount

Given:

In how much time would the simple interest on a principal amount be 0.125 times the principal amount at 10% per annum

Formula used:

SI = P × R × T / 100

Where,

SI = simple interest

P = principal

R = rate

T = time

Calculation:

Let the principal be P and the time be T years

According to the question

⇒ 0.125 × P = SI

⇒ 0.125P = P × 10 × T / 100

⇒ 0.125P = P × T / 10

⇒ T = 10 × 0.125P / P 

⇒ T = 10 × 0.125 = 1.25

⇒ T = 125 / 100

⇒ T = 5 /4

∴ The time annum is 5 / 4 years

Let B and C can do a work in b and c days respectively.

Given that,

And,

Again,

Given:

L.C.M. and product of any two positive numbers are 245 and 980 respectively

Formula Used:

Product of two number ( a × b) = L.C.M. of two number(a, b) × H.C.F. of two number(a, b)

Calculation:

980 = H.C.F × 245

⇒ H.C.F. = 980/245

⇒ H.C.F. = 4

∴ The H.C.F. of these number is 4.

Given :
104.96 + 120.96 + 103.16 - 12.89 × 2.04 + 124.93 ÷ 5.1 = ?
Formula used:

Calculation:
104.96 + 120.96 + 103.16 − 12.89 × 2.04 + 124.93 ÷ 5.1
= 104.96 + 120.96 + 103.16 − 12.89 × 2.04 + 24.5
= 104.96 + 120.96 + 103.16 − 26.3 + 24.5
= 353.58 − 26.3
= 327.73 ∼ 328
Hence the closest approximate value is 328.

Given:

The average age of A and B is 30 years. that of B and C is 32 years and that of C and A is 34 years.

Concept used:

Average

Calculation:

The average of A and B is 30 years

⇒ (a + b) / 2 = 30

⇒ (a + b) : 30 × 2 = 60

The average of B and C is 32 years

⇒ (b + c) / 2 = 32

⇒ (b + c) = 32 × 2 = 64

The average of A and C is 34 years

⇒ (a + c) / 2 = 34

⇒ (a + c) = 34 × 2 = 68

Average : (60 + 64 + 68)/2 = 96 years

(a + b + c) = 96

Age of c = 96 - (a + b)

Age of c = 96 - 60

Age of c = 36 years.

The answer is 36 years.

Given:

Distance in still water = 135 km

Time taken is still water = 7.5 hours

Distance in upstream = 48 km

Time is taken in upstream = 4 hours

Formula used:

Speed = Distance/Time

Calculations:

⇒ Speed of a boat in still water = 135/7.5

⇒ Speed of a boat in still water = 18 km/hr

Let, the speed of the current is x.

⇒ Speed of a boat upstream = (18 - x) km/hr

⇒ Speed in upstream = (18 - x) = 48/4

⇒ (18 - x) = 12

⇒ x = 6 km/hr

∴ The speed of current is 6 km/hr.

Let quantity of milk and water at 10:00 AM is ‘x + 90’ and ‘x’ respectively.
Quantity of water vaporized from 10:00 AM to 4:00 PM = 2.5 × 6 = 15 liters
According to the question:
x + 90/ x - 15 = 8/1
x + 90 = 8x – 120
7x = 210
x = 30
Quantity of milk at 8:00 PM = (x + 90) = 120 liters
Quantity of water at 8:00 PM = x – 2.5 × 10 = 30 – 25 = 5 liters
Required percent = (120/125) × 100 = 96%

Formula:
Minimum number of seats required = LCM of 4, 6, or 10
Concept:
To find the minimum number of seats, we need to find the LCM of the number of seats that groups should contain. The LCM of the number of seats would give a least common multiple i.e., the number of seats that could be fully utilized while dividing the class into respective groups.
Calculation:
LCM of 4, 6, or 10 = 60
∴ Minimum number of seats required = 60
The correct answer is option (B)

Calculation:
Let assume that the natural number is 18
⇒ 18 ÷ 5
Q = 3, R = 3
⇒ 18 ÷ 2
Q = 9, R = 0
⇒ 18 ÷ 10
Q = 1, R = 8
This statement is false.
The correct option is A i.e. False

Given:

July, August and September income = Rs. 45,000; Rs. 65,000 and Rs. 7,000

Average expenses = Rs. 24,000

Formula used∶

Average = Sum of all observations / Number of observations

Calculation∶

Total income of July, August and September = Rs. (45,000 + 65,000 + 7,000)

⇒ Rs. 1,17,000

Total expenses of July, August and September = 24,000 × 3

⇒ Rs. 72,000

Saving of July, August and September = Rs. (1,17,000 - 72,000)

⇒ Rs. 45,000

Average saving = 45,000 / 3

⇒ Rs. 15,000

∴ The average saving of these months is Rs. 15,000.

Given:

Calculation:

The number of Washing Machine sold during the quarterly period of winter concession in all cities

⇒ 522 + 419 + 715 + 822

⇒ 2478

The number of Washing Machine sold during the quarterly period of millennium concession in all cities

⇒ 468 + 452 + 735 + 814

⇒ 2469

According to the question,

⇒ 2478 – 2469

⇒ 9

∴ 9 more than those of the millennium concession period.

Given:

The simple interest = Rs. 3696

Rate of interest = 50%

Time duration = 8 year

Formula used:

Interest = PTR/100

where P = Principal or the sum lent, R = Rate of interest, T = Time duration and I = Simple interest

Calculation:

The simple interest = Rs. 3696

⇒ I = Rs. 3696

Rate of interest = 50%

⇒ R = 50%

Time duration = 8 year

⇒ T = 8 year

Interest = PTR/100

⇒ 3696 = (P × 50 × 8)/100

⇒ P = 924

∴ The principal is Rs. 924.

Given:

• Total tourist traffic is 20 lakhs.
• No. of tourists visiting from UP, Haryana, Rajasthan, and others are 10%, 40%, 30%, and 20% respectively.

Concept used:

• No. of tourists= percentage× total tourist traffic

Calculations:
We know that,

• No. of tourists from Haryana = percentage × total tourist traffic = 40% × 20 lakh = (40 × 20)/100 = 8 lakhs
• No. of tourists from Rajasthan = percentage × total tourist traffic = 30% × 20 lakh = (30 × 20)/100 = 6 lakhs

Difference between the no. of tourists visiting from Haryana and Rajasthan= No. of tourists from Haryana-No. of tourists from Rajasthan = 8 - 6 = 2 lakhs
Therefore, the correct answer is 'option C'.

Concept:

A number is divisible by 2 if its last digit is 0, 2, 4, 6, or 8.

A number is divisible by 3 if the sum of its digits is divisible by 3.

A number is divisible by 11 if the difference between the sum of digits occupying even and odd places is divisible by 11 (or is zero).

Calculations:

A) 8448:
Visibly, it's divisible by 2 (8 is an even number).
The sum of its digits is 8 + 4 + 4 + 8 = 24, which is divisible by 3.
The difference between the sum of digits at even places (4 + 8 = 12) and at odd places (8 + 4 = 12) is 0, which is divisible by 11. So, 8448 is divisible by 2, 3 and 11.

B) 9812:
Visibly, it's divisible by 2 (2 is an even number).
The sum of its digits is 9 + 8 + 1 + 2 = 20, which is NOT divisible by 3.
So, we don't need to check for divisibility by 11. 9812 is NOT divisible by 2, 3 and 11.

C) 9126:
Visibly, it's divisible by 2 (6 is an even number).
The sum of its digits is 9 + 1 + 2 + 6 = 18, which is divisible by 3.
The difference between the sum of digits at even places (1 + 6 = 7) and at odd places (9 + 2 = 11) is 4, which is NOT divisible by 11. So, 9126 is NOT divisible by 2, 3 and 11.

D) 9636:
Visibly, it's divisible by 2 (6 is an even number).
The sum of its digits is 9 + 6 + 3 + 6 = 24, which is divisible by 3.
The difference between the sum of digits at even places (6 + 6 = 12) and at odd places (9 + 3 = 12) is 0, which is divisible by 11. So, 9636 is divisible by 2, 3 and 11.

So, among the options C i.e A & D is Correct.

According to the question,

Let the period Nishant remained in business be n

Ratio of investment = 25000 × 12 : 50000 × n = 6 : n

Given that profit was distributed in the ratio 3 : 1

⇒ 6/n = 3/1

⇒ n = 2

Nishant joined the business after (12 – 2) = 10 months