RRB NTPC Mathematics Test - 4 - RRB NTPC/ASM/CA/TA MCQ

Given:

Height of a cone = 12 cm

Radius of the cone = 3.5 cm

Formula used:

Volume of the cone = 1/3 πr²h where,

Height = h

Radius = r

Calculation:

According to the question,

Volume of the cone = 1/3 × 22/7 × (3.5)² × 12

⇒ 1/3 × 22/7 × 3.5 × 3.5 × 12

⇒ 154 cm³

∴ The volume of the cone is 154 cm³.

Given data:
Cost of solution = Rs. 16 per litre,
Cost of Acetone = Rs. 14 per litre,
Ratio of Toluene to Acetone in the solution = 4 ∶ 1.
Concept used:
The cost of the solution is the sum of the cost of its components in their respective ratios.
Represent the cost of solution as a weighted average of the costs of Toluene (T) and Acetone (A):
⇒ 16 = (4T + 1 × 14) / 5
Rearrange to solve for T (Cost of Toluene):
⇒ T = (80 - 14) / 4
Therefore, substituting the values from step 2 into the equation:
T = (80 - 14) / 4 = 66 / 4 = Rs. 16.5/litre
Hence, the cost of Toluene per litre to the scientist is Rs. 16.5.

Given:

The mixture of x and y in Alloy A = 5 : 2

The mixture of x and y in Alloy B = 3 : 4

The ratio of A and B in alloy C = 4 : 5

Calculation:

Let the quantity of metal x in alloy C be x

Quantity of metal x in alloy A = 5/7

Quantity of metal y in alloy A = 2/7

Quantity of metal x in alloy B = 3/7

Quantity of metal y in alloy B = 4/7

According to the question

The ratio of x and y in alloy C = [(5/7 × 4) + (3/7 × 5)]/[(2/7 × 4) + (4/7 × 5)]

⇒ (20/7 + 15/7)/(8/7 + 20/7)

⇒ (35/7)/(28/7)

⇒ (35/7 × 7/28)

⇒ 5/4

Now,

Quantity of x in alloy C = 5/(5 + 4)

⇒ 5/9

Percentage of x in alloy C = (5/9 × 100)

⇒ 500/9

⇒

∴ The required percentage of x in alloy C is

Given:

Cost of 90 articles = Rs 5400

Formula used:

Gain = SP – CP

Loss = CP – SP

Loss % = (Loss/CP) × 100

Gain % = (Gain/CP) × 100

Calculation:

Cost of 1 article = 5400/90 = Rs. 60

Total cost price of 70 articles = 70 × 60 = Rs. 4200

Profit earned = 5040 – 4200 = Rs. 840

% profit = (840/4200) × 100 = 20%

∴ There is a gain of 20%.

According to the question:
Average = (12000 + 10000 + 5000 + 20000 + 15000)/5 = 12,400
∴ Option B is the correct answer.

Given,

Due to engine trouble, an Express train goes at 9/10^th of its usual speed and arrives at 2 : 34 pm instead of 2 : 28 pm

Concept:

Speed ratio of two train = x : y

Time ratio of two train = y : x

As we know, time is inversely proportional to speed.

Calculation:

Difference in time = 2:34 – 2:28 = 0:06 min

Let speed of express train be 10x

⇒ Speed of train after trouble in engine = 10x × 9/10 = 9x

Speed ratio of train = 10x : 9x

⇒ Time ratio of train = 9x : 10x

According to the question,

10x – 9x = 6 min

⇒ x = 6 min

⇒ Original time = 10 × 6 = 60 min = 1 hour

∴ The correct time when the train start = 2:34 – 1:00 = 1:34 PM

Given:

Divisor = 42

Quotient = 120

Remainder = 0

Concept:

The number divided can be represented as divisor × quotient + remainder.

Number = Divisor × Quotient + Remainder

⇒ Number = 42 × 120 + 0

⇒ Number = 5040

Now, when this number is divided by 41,

⇒ 5040/41

⇒ Remainder = 38

Therefore, the remainder when the same number is divided by 41 is 38.

Given:

The marked price of a cap is ₹2,000. Two successive discounts of 15% and x% are given and the selling price is ₹1,275.
Formula used:

Discounted price = × marked price
Calculation:

With successive discounts of 15% and x% in 2000,

Solving for x,

∴ The value of x is 25.

Given :
Secθ = 4/3
Calculation :
We know,
⇒ Secθ = 4/3, tanθ = √7/3
⇒ tan² θ + tan⁴ θ = 7/9 + 49/81
⇒ (63 + 49)/81
⇒ 112/81
∴ The correct answer is 112/81.

Given,

Pipe A can fill a tank in X hours

Pipe B can empty the full tank in 15 hrs

If both pipe A and pipe B open, then the tank filled in 7 hrs 30 min = (7 + 1/2) = 15/2 hrs

Formula:

Total work = Efficiency × Time taken

Calculation:

Total work = 15 units

⇒ Efficiency of B = (-1) units/hr

Let efficiency of A be t, then

t – 1 = 2

⇒ t = 2 + 1

⇒ t = 3 units/hr

∴ Pipe A alone can fill the tank in = 15/3 = 5 hrs

Given:

Concept used:

Calculation:

⇒

⇒

⇒

⇒

⇒

⇒ = 2

∴ The simplified value is 2.

Given:

ST || QR and ST : QR = 3 : 8.

Formula Used:

Ratio of area of similar triangle = Ratio of square of corresponding sides

Calculation:

We know that,

∴ Area of ΔPST = 9

Area of trapezium QRTS = Area of ΔPQR - Area of ΔPST

Area of trapezium QRTS = 64 - 9 = 55.

∴ The ratio of the area of ΔPST and trapezium QRTS is 9 : 55.

Given:

The speed of the train = 29 m/s

Time to pass a station platform = 42 seconds

Time to pass a man = 26 seconds

Formula used:

Speed of the train = Length of (train + platform)/Time to pass a station platform

Speed of the train = Length of train/Time taken to pass a man

Calculation:

Speed of the train = Length of train/Time taken to pass a man

⇒ 29 = Length of train/26

⇒ Length of train = 29 × 26 = 754 m

Speed of the train = Length of (train + platform)/Time to pass a station platform

⇒ 29 = (754 + Length of platform)/42

⇒ 754 + Length of platform = 29 × 42

⇒ 754 + Length of platform = 1218

⇒ Length of platform = 1218 - 754 = 464 m

∴ The length of the platform (meter) = 464 m

Given:

Principal, P = 3000

Rate of Interest, R = 5%

Number of years, N = 2

Formula used:

Amount, A = P[1 + (R/100)]^N

C.I = A - P

Calculation:

A = P[1 + (R/100)]^N

⇒ A = 3000[1 + (5/100)]²

⇒ A = 3307.5

C.I = A - P

⇒ C.I = 3307.5 - 3000

∴ C.I = Rs. 307.5

Given:

Father = 4 × daughter

Five years later

Father = 3 × daughter.

Calculation:

Let the present age of daughter be x years.

So, the father's present age is 4x years

After 5 years

⇒ 4x + 5 = 3 × (x + 5)

⇒ 4x + 5 = 3x + 15

⇒ x = 10

Present age of daughter is 10 years

Father's present age is 4 × 10 = 40 years

Ratio asked after 10 years

Age after 10 years will be

⇒ Father : Daughter = 50 : 20

⇒ Father : Daughter = 5 : 2

∴ Father's age will be 2.5 times of daughter age after 10 years.

⇒ Father : Daughter = 4 : 1

⇒ (Father + 5)/(Daughter + 5) = 3/1

⇒ (4x + 5)/(x + 5) = 3/1

⇒ 4x + 5 = 3x + 15

⇒ x = 10

Father = 40 years and daughter = 10 years

Ratio after 10 years will be

⇒ 50 : 20 = 5 : 2

∴ Father's age will be 2.5 times of daughter age after 10 years.

Shortcut Trick

In the given ratio we need to make the difference of ratio equal

After equating difference between the father's age is equal to given years

so 1unit = 5 years

In the question asked 5 years more later so wee add 1 unit into the ratio

Final ratio we will get Father : Daughter = 10 : 4

∴ Father's age will be 2.5 times of daughter age after 10 years.

Using the rule of allegation, if two quantities are mixed, then

⇒ Quantity of cheaper/Quantity of dearer = (Cost price of Dearer – Mean Price)/(Mean Price – Cost price of cheaper)

⇒ Substituting the values, Quantity of cheaper/Quantity of dearer = (31 – 24.50)/(24.50 – 20)

⇒ Quantity of cheaper/Quantity of dearer = 6.5/4.5

⇒ Ratio is 6.5 : 4.5

In the figure, AB is the tower and CD is the distance between two men on either side of tower;

In ΔABD

tan 60° = AB/AD

⇒ √3 = 75/AD

⇒ AD = 75/√3

tan 30° = AB/AC

⇒ 1/√3 = 75/AC

⇒ AC = 75√3

∴ CD = 75√3 + 75/√3 = 75√3 + 25√3 = 100√3

∴ Distance between two men on either side of tower = 100√3 m

We know that,

Cosecθ = 1/sin θ ⇒ sin θcosecθ = 1, and cotθ = 1/tanθ ⇒ cotθ tanθ = 1

∠OPT = 90° [∵ Radius is perpendicular to tangent]

∠OPQ = 90° - ∠QPT = 90° - α

∠OQP = 90° - α [∵ OQ = OP]

In triangle OQP

∠O + ∠Q + ∠P = 180°

∠O + 90 - α + 90 - α = 180

∴ ∠O = 2α

Calculation:

Total expenditure = 48 cr

The total expenditure of the company Q, R & T together = (22 + 18 + 20)% = 60%

Now, according to the question

The total expenditure of the company Q, R & T together = 60% of total expenditure

⇒ (60% of 48) = ((60/100) × 48)

⇒ 144/5 = 28.8

∴ The required expenditure is 28.8

Given:

Principal = Rs. 2106

Time = 3 years

Rate = 8%

Formula used:

Annual instalment =

Calculation:

Annual instalment =

=

=

=

= 650

Answer is 650.

Given:

An equilateral triangle ABC

AB = BC = CA = 12 cm

Concept used:

Centroid is divided a median in the ratio of 2 : 1

Length of median of equilateral triangle = √3a/2

where a is the side of equilateral trianlge

Calculation:

According to the concept used,

Median AD = BE = CF = √3a/2

⇒ AD = (√3/2) × AB

⇒ AD = √3 × 12/2

⇒ AD = 6√3

Now, again according to the concept used

⇒ AG = (2/3) × AD

⇒ AG = 4√3

∴ AG is 4√3

Shortcut Trick

In an equilateral trianlge ABC, If 'G' is the centroid, then

AG = AB/√3

AG = 12/√3

∴ AG is 4√3

Given:
Largest right circular cone that can be cut out of a cube where the edge is 9 cm
Formula used:
Volume of cone = 1/3 × π × radius × radius × height
Calculations:

The base of the largest right circular cone will be the circle inscribed in the face of the cube and its height will be equal to an edge of the cube.
Radius of the base of the cone r = and height = 9
Radius = height/2 = 9/2
Required volume = 1/3 × 22/7 × 9/2 × 9/2 × 9
=> 2673/14 = 190.93 ~191 cm³
∴ The answer is 191 cm³.

Let the cost price of 1 kg sugar be Rs. 100

He takes 1.25 kg at the cost of 1 kg, then cost price of 1 kg of sugar for shopkeeper = 100/1.25 = Rs. 80

He gives 0.8 kg at the cost of 1 kg, then selling price of 1 kg of sugar for shopkeeper = 100/0.8 = Rs. 125

Profit amount = 125 - 80 = Rs. 45

Given∶

The average weight of 20 players = 40 kg

The average weight increased by 500 g if weight of coach is added.

Formula Used∶

Sum of observation = Average × Number of observations

Calculation∶

Average weight of 20 players = 40 kg

Total weight of 20 players = 40 × 20 = 800 kg

Average weight of 20 players and the coach = (40 + 0.5) kg = 40.5 kg

Total weight of 20 players and the coach = (40.5 × 21) kg = 850.5 kg

So, weight of the coach = 850.5 kg - 800 kg = 50.5 kg

Cost price = Rs. 1225

Since he earns a profit of 20%;

∴ Selling price = 1225 × 1.2 = Rs. 1470

Since a discount of 12.5% is allowed on the mark price;

As a = b ⇒ a³ – b³ = 0

Substitute a = b
0 + a/a - a
So, the expression gets simplified to:
1 – a

Given:

We have to evalute 2 tan² 45° + cos² 30° - sin² 60°

Formula Used:

tan45° = 1

Cos30° = √3/2

Sin60° = √3/2

Calculation:

2 tan² 45° + cos² 30° - sin² 60°

⇒ 2 × 1² + (√3/2)² – (√3/2)²

⇒ 2 + 3/4 – 3/4

⇒ 2

∴ The value of 2 tan² 45° + cos² 30° - sin² 60° is 2.

Given:

The surface area of the sphere = 616 cm²

Formula Used:

Surface area of sphere = 4 × πr²

Volume of sphere = (4/3) × πr³

Calculation:

Surface area of sphere = 4 × πr²

⇒ 616 = 4 × (22/7) × r²

⇒ r² = 616 × 7/88

⇒ r² = 7 × 7

⇒ r = 7

TSA of one of the spheres = 2πr² + πr²

⇒ 3πr²

⇒ 3 × (22/7) × (7)²

⇒ 3 × 22 × 7

⇒ 462

∴ The total surface area of one of the hemispheres is 462 cm².

Given:
In ΔABC, ∠C = 90°, BC = AC = 3√2,
Calculation:
In ΔABC, ∠C = 90°, BC = AC = 3√2,
By Pythagoras theorem,
AB² = AC² + BC²
⇒ AB² = ( 3√2)² + ( 3√2)² = 18 +18 = 36
AB = 6 cm