RRB NTPC Mathematics Test - 5 - RRB NTPC/ASM/CA/TA MCQ

Formula:

P =

CI = A - P

where-

P = Principal, SI = Simple Interest

R = Rate, T = Time

CI = Compound Interest, A = Amount

Calculation:

Using the SI formula, we get-

Amount accrued to Anil, when invested at compound Interest,

Total amount = Rs 15,680

CI = Rs 15,680 - Rs 12,500 = Rs 3,180

∴ The compound interest on the sum is Rs 3,180.

Reduce the following fraction to its lowest number
Option A
9/15 = 3/5 (common multiple = 3)
Option B
9/20 = 9/20 (no common multiple)
Option C
9/25 = 9/25 (no common multiple)
Option D
3/20 = 3/20(no common multiple)
∴ Option A is the wright answer

Given:

The marked price of an item is Rs. 15,800.

The shopkeeper earns a profit of 18% after allowing a discount of 20%.

Concept used:

Selling price = Marked Price - Marked Price × Discount%

Selling Price = Cost Price + Cost Price × Gain%

Calculation:

Selling price of the item = 15800 - 15800 × 20% = Rs. 12640

Let the cost price of the item be P.

According to the question,

P + P × 18% = 12640

⇒ 1.18P = 12640

⇒ P = 12640/1.18

⇒ P = 10711.86

⇒ P ≈ 10712

∴ The cost price of the item is Rs. 10,712.

Given:

Rahul can complete the one round in = 15 minutes

Anil can complete the one round in = 18 minutes

Concept:

The LCM is the smallest multiple of two or more than two numbers.

Calculation:

Let the first meeting point time is X

According to the question,

LCM of (15, 18) = X

⇒ 15 = 3 × 5

⇒ 18 = 2 × 3 × 3

The LCM of (15 and 18) = 2 × 3² × 5 = 90

∴ The required result will be 90.

Given:

Initial investment by A = ₹75,000

B joins 5 months later with an investment = ₹80,000

Total profit after 1 year = ₹4,08,800

Concept Used:

Profit shares ∝ (Investment × Time Period)

Calculations:

A's Capital-Time = ₹75,000 × 12 months = ₹9,00,000 months

B's Capital-Time = ₹80,000 × 7 months = ₹5,60,000 months

Ratio of A's and B's capital-time:

⇒ A : B = ₹9,00,000 : ₹5,60,000

⇒ A : B = 90 : 56

⇒ A : B = 45 : 28

Share of total profit according to ratio:

⇒ A's share = (45/(45 + 28)) × ₹4,08,800 = ₹2,52,000

⇒ B's share = (28/(45 + 28)) × ₹4,08,800 = ₹1,56,800

∴ A's share = ₹2,52,000 and B's share = ₹1,56,800.

Formula :
SP = MP × (1 - D%)
Where, MP = Marked Price, D% = Discount Percentage
Calculation:
Let the cost price be 100 units
⇒ MP = (1 + 45/100) × 100 = 145 units
According to the question,
SP = 145 × (1 - 40/100) = 87 units
∴ Net decrease in the price = 87 - 100 = -13 units
So, In percentage = (13units /100 units) × 100 = 13%
∴ The shopkeeper gets Rs 13 less on the decided price.
Additional Information
The negative sign in the answer is for representational purposes. The presence of the negative sign shows that the shopkeeper sold the product at a cost lesser than the original cost i.e., incurred a loss.

Given:
The length of the one diagonal of a rhombus is 30 cm.
The area of the rhombus is 300 cm².
Concept:
The area of a rhombus with diagonals as D₁ and D₂ is 1/2 x D₁ x D₂ square units.
As per the question,
The length of the one diagonal of a rhombus is 30 cm.
Let us assume the other diagonal is D₂ cm.
Now,
The area of the rhombus is 300 cm² = 1/2 x D₁ x D₂
1/2 × 30 × D₂ = 300
⇒ D₂ = 20 cm
Hence, option C is correct.

Given:

Total number of students is 600000.

Calculation:

Out of 600000, 40% passed, the total number of passed students 600000 × 40/100 = 240000

Out of 600000, 40% were female, the total number of females = 240000 and males = 360000

The pass percentage among the males was 50%, total males passed= 360000/2 = 180000

So, female passed = (240000 - 180000) = 60000

So, female pass% = 60000/240000 × 100 = 25%

∴ The correct answer is 25%

Given:

Winning candidate gets 70% of votes.

The candidate wins by a majority of 500 votes.

Concept Used:

The Total number of votes = Votes of winning candidate + Votes of losing candidate.

Calculation:

Let the total number of votes be x.

Votes for winning candidate = 70% of x = 0.70x

Votes for losing candidate = 30% of x = 0.30x (since the total votes have to be 100%)

The Total number of votes = Votes of winning candidate + Votes of losing candidate.

⇒ 0.70x - 0.30x = 500

⇒ 0.40x = 500

⇒ x = 500 / 0.40 = 1250

Therefore, the total number of votes is 1250.

Given:

Principal = Rs. 24,000

Amount = Rs. 26,460

Rate of interest = 5% p.a.

Concept Used:

The formula for compound interest is: A = P (1 + r/n)^(t), where n is the number of times interest is compounded per time period t.

⇒ 26,460 = 24,000 (1 + 5/100)^(t)

⇒ (26,460/24,000) = (21/20)^(t)

⇒ t = log[(26,460/24,000)] / log[1.05] ≈ 2 years

⇒ In months, t = 2 x 12 = 24 months

Therefore, it will take 24 months for the sum to amount to Rs. 26,460.

Given:

A can paint in = 50days,

B can paint in = 10days,

A + B + C can paint in = 6.25days

Calculation:

Total work = LCM of {50, 10, 6.25} = 50

Efficiency of A = 50/50 = 1 unit

Efficiency of B = 50/10 = 5 unit

Efficiency of A + B + C = 50/6.25 = 8 unit

Now, Efficiency of A + B = (1 + 5) units= 6 units

So, Efficiency of C = 8 - 6 = 2 units

Thus,

C can also paint the wall = 50/2 = 25 days.

Given:

Quantity of mixture = 5 litre

Percentage of water in the mixture = 13%

Quantity of water added to the mixture = 1 litre

Calculation:

Amount of water present in the mixture = 13% of 5 litre = 650 ml water

After adding Water to the mixture, water in the mixture = (1000 + 650) = 1650 ml

Quantity of milk present in 6 litre of mixture = (6000 - 1650) = 4350 ml

Ration of Milk and water in the mixture = 4350 : 1650 = 174 : 66 = 87 : 33

Calculation:

Sales for 40^'' size LCD = 277

Sales for 50'' size LCD = 238

Sales for 55'' size LCD = 245

Sales for 60'' size LCD = 247

Sales for 65'' size LCD = 111

Sales for 70'' size LCD = 84

Sales for 75'' size LCD = 62

∴ The correct answer is 75^''.

Given:

BC = 18 m

Concept:

Formulae Used:

Tanθ = Perpendicular/Base

Cosθ = Base/Hypotenuse

Calculation:

Height of the tree = AB + AC

Tan 30° = AB/18

⇒ (1/√3) = AB/18

⇒ AB = (18/√3)

Cos 30° = BC/AC = 18/AC

⇒ √3/2 = 18/AC

⇒ AC = 36/√3

Hence, AB + AC = 18/√3 + 36/√3 = 54 / √3

⇒ 54/√3 × √3 /√3 (rationalizing to remove root from denominator)

⇒ 54√3 / 3 = 18√3

∴ Height of the tree = 18√3.

Mistake Point:
Here, total height of tree is (AB + AC).
The above Question is previous year Question taken directly from NCERT class 10th. Correct answer will be 18√3

Given∶

A right angled triangle and a circle is inscribed in it.

The lengths of the sides containing the right angle are 12 m and 16m.

Formula Used∶

Area of triangle = 1/2 × Base × Height

Use of Pythagoras theorem

Calculation∶

In ΔBAC right angled at A.

AB = 12 m and AC = 16 m

Using Pythagoras theorem, we have

BC² = AB² + AC²

⇒ BC² = 12² + 16² = 144 + 256 = 400

⇒ BC = 20

Now, Circle is inscribed in the triangle such that if touches the sides of a triangle at point D, E and F. Join the center O with D, E and F.

Then, OD = OE = OF = r.

Also, OD, OE and OF are perpendicular on AB, AC and BC.

Area of Δ ABC = Area of ΔOAB + Area of Δ OBC + Area of Δ OCA.

⇒ 1/2 AB × AC = 1/2 AB × r + 1/2 × BC × r + 1/2 × CA × r

⇒ 1/2 × 12 × 16 = 1/2 (12 × r) + 1/2 (20 × r) + 1/2 (16 × r)

⇒ 192 = 12r + 20r + 16r

⇒ 192 = 48r

⇒ r = 4 m

∴ The radius of the circular door is of 4 m.

Given:

a + b = 20 and ab = 75

Formula used:

a³ + b³ = (a + b)(a² + b² – ab)

a² + b² = (a + b)² – 2ab

Calculation:

a³ + b³ = (a + b)(a² + b² – ab)

⇒ (a + b)((a + b)² – 3ab)

⇒ 20(20²– 3 × 75)

⇒ 20 × 175

∴ The value of a³ + b³ is 3500

Shortcut Trick
Factorization of 75 = 5 × 5 × 3
Here the value of a and b should be 5 and 15 or 15 and 5 which satisfies ab = 75
⇒ a³ + b³ = 5³ + 15³ = 15³ + 5³ = 3500
∴ The value of a3 + b3 is 3500

Given:

P1 = 6 hours

P2 = 8 hours

P3 = 16hours

After 3 hours P3 is opened.

Concept used:

Time taken = 1/Efficiency

LCM(6, 8, 16) = 48

The total capacity of the tank = 48 units

The capacity of P1 to fill the tank in 1 hour = 48/6 = 8 units

Capacity of P2 in 1 hour = 48/8 = 6 units

The capacity of P1 and P2 in 3 hours = 3 x (6 + 8) = 42 units

Capacity of P3 = 48/16 = 3 units

Capacity of P1, P2, P3 = 6 + 8 - 3 = 11 units

Tank empty after 3 hours = 48 - 42 = 6 units

Time required to fill the left empty tank = 6/11 hours

Total time to fill the tank = 3 + 6/11 = 39/11 hours

Hence, the correct option is D.

Given

Principal = Rs 8000

Rate of interest = 10%

Formula used

Amount = Principal(1 + R/100)^t

Calculation

For semi annually, rate of interest = 10%/2 = 5%

⇒ 9261 = 8000(1 + 5/100)^t

⇒ (21/20)³ = (21/20)^t

⇒ t = 3

⇒ No. of years = t/2 = 3/2 (As compounded semi annually)

∴ Amount will become Rs 9261 is (3/2) years

In ΔTOP, tan θ = OT/OP

• tan θ = OT/ m²

In ΔTOQ, tan(90 - θ) = OT/OQ

• tan (90 - θ) = OT/ n² 

(1) x (2) by multiplying the equations 1 & 2

tan θ x cot θ = (OT/ m² ) x (OT/ n²)

m² n² = (OT)²
OT = mn.

Given:

⇒

⇒ 3 + k = 5

⇒ k = 2

∴ Option B is the correct answer.

Total people in meeting is 45 and out of then 40 people know each other.

So 5 people don't know anyone.

Let those 5 peoples be A, B, C, D, E

So A will handshake 44 people.

B will handshake 43 people

C will handshake 42 people

D will handshake 41 people

and E will handshake 40 people

Hence total handshake = 44 + 43 + 42 + 41 + 40 = 210

Option (C) is correct

∠PSR = 105°

So, angle ∠RQP = 75°

PQ is diameter so angle PRQ = 90°

∠RPQ = 180 − (90 + 75)

∠RPQ = 15°

Calculation:

In the month of December 31 day, we have 31 days i.e.

28 + 3 days

for 28 days there would be 4 Sundays.

for the remaining 3 days we need to consider.

if we consider the first 3 days of the month of December, and to get 5 Sundays in the month of December, Sunday should be in the first three days.

i.e, it could be

1^st December Sunday

or 2^nd December be Sunday

or 3^rd December be Sunday.

We know that the Total days in a week = 7

The probability of getting 1 December be Sunday = 1/7

Similarly, 2^nd and 3^rd to be Sunday = 1/7 and 1/7 respectively.

The total probability of getting 5 Sundays in the month of December would be

= 1/7 + 1/7 + 1/7

= 3/7

or

if we consider the last 3 days

i.e., 28 days + 3 days

The possible outcome for the last 3 days would be

• The last three days fall on Monday, Tuesday, and Wednesday.
• The last three days fall on Tuesday, Wednesday, and Thursday.
• The last three days fall on Wednesday, Thursday, and Friday.
• The last three days fall on Thursday, Friday, and Saturday.
• The last three days fall on Friday, Saturday, and Sunday.
• The last three days fall on Saturday, Sunday, and Monday.
• The last three days fall on Sunday, Monday, and Tuesday.

There are a total of 7 outcomes out of which Sundays will fall on
Friday, Saturday, and Sunday.
Saturday, Sunday, and Monday.
Sunday, Monday, and Tuesday.
So, the probability of getting 5 Sundays = 3/7
The correct answer is option (C)

Fourth proportional∶

(x + 3) ∶ 8 ∶∶ 27 ∶ (x – 3)

⇒ (x + 3) ∶ 8 = 27 ∶ (x – 3)

⇒ (x + 3)/8 = 27/(x – 3)

⇒ (x + 3)(x – 3) = 27 × 8

⇒ x² – 9 = 216

Given:

5 years ago, A man was 7 times = As his son

Now in 5 years, A man will be = 3 times as old as his son

Calculation:

Let five years ago, son age be x and father age be 7x respectively

Present age of son = (x + 5)

Present age of father = (7x + 5)

5 years later, son age = (x + 5 + 5) = (x + 10)

5 years later, father age = (7x + 5 + 5) = (7x + 10)

According to the question

⇒ (7x + 10) = 3(x + 10)

⇒ 7x + 10 = 3x + 30

⇒ 7x – 3x = 30 – 10

⇒ 4x = 20

⇒ x = (20/4)

⇒ x = 5

Present age of son = (x + 5) = (5 + 5)

⇒ 10

Present age of father = (7x + 10) = (7 × 5 + 5)

⇒ 40

Now,

Son age two years ago = (10 – 2) = 8 years

Father age two years ago = (40 – 2) = 38 years

∴ The required age is 38 years and 8 years

Calculation

Let the marks scored in first semi ster is x

Marks scored in second semester = 112x/100

Marks scored in third semester= (112x/100) - (0.1344x)

= 1.12x - 0.1344x

= .9856x

Percentage decrease = (0.0144x/x) × 100

= 1.44%

Given:

Let the CP of article = Rs. 100

So, MP of Article = 125% of Rs. 100 = Rs. 125

Formula used:

MP/CP = (100 + Profit%)/(100 - Discount%)

Calculations:

125/100 = (100 + P%) / (100 - 10)

5/4 = (100 + P%) / (90)

(100 + P%) = (90) × (5/4)

(100 + P%) = 112.5

P% = 12.5%

Profit% = 12.5%

Alternate Method

Given:

Marked price of a articles is 25% more than cost price.

Discount given on the article is 10%.

Formula used:

Discount% = (Discount/M.P) × 100

Profit% = (Profit/C.P) × 100

S.P = M.P – Discount

Profit = S.P – C.P

Calculation:

Let the cost price of article be 'x'.

So, Marked price of article = x + 25% of x

⇒ (5x)/4

10% of Marked price = (10/100) × (5x/4) = x/8

Selling Price = M.P – Discount

⇒ (5x)/4 – (x/8)

⇒ (9x/8)

Profit = S.P – C.P

⇒ (9x/8) – x

⇒ x/8

Profit% = (Profit/C.P) × 100

⇒ Profit% = [(x/8)/x] × 100

⇒ 100/8

⇒12.5%

∴ The profit percentage on the article is 12.5%.

Given:
For a glass of Lassi, he uses 5/6 of curd and rest is milk.
Formula Used:
F = I (1 - X/V), where
F = fraction of curd in final mixture
I = fraction of curd in initial mixture
X = Quantity of mixture replaced
V = Total volume of mixture.
Calculation:
Use the formula, F = I ( 1 - X/V), where
F = 1/2, I = 5/6, X/V = ?
Substitute the values in above formula
1/2 = (5/6) ( 1- X/V)
⇒ 1/2 × 6/5 = 1 - X/V
⇒ X/V = 1 - 3/5 = 2/5
∴ The required ratio is 2 : 5.

Given:

Ramesh invested Rs. 100000 at compound interest and same amount at simple interest in two different banks both giving 5% rate of interest per annum.

Concept:

For CI:

Cumulative rate for x% and y% = (x + y + xy/100)

Calculation:

We can calculate the difference of cumulative rate of interests in case of CI and SI:

So,

Cumulative rate of interest for SI = 5 + 5 = 10%

And

Cumulative rate of interest for CI = 5 + 5 + (25/100) = 10.25%

Hence,

Difference of cumulative rate of interests = 10.25 – 10 = 0.25%

∴ CI – SI = 100000 × (0.25/100) = Rs. 250

Shortcut Trick

CI – SI = P x (R/100)² = 100000 x (5/100) x (5/100) = Rs. 250

So, AB is a wall & AC is ladder of constant length x

Let’s say AB = y

Given that BC = 2.5

Then by Pythagoras theorem

y² + 2.5² = x²

y² + 6.25 = x² ----(1)

Now, ladder slips by 0.8 m & bottom away from the wall’s 1.4 m

Then

(y - 0.8)² + (2.5 + 1.4)² = x²

(y - 0.8)² + (3.9)² = x² ----(2)

y² - 1.6y + 0.64 + 15.21 = x²

y² - 1.6y + 15.85 = x²

y² - 1.6y + 15.85 = y² + 6.25

1.6y = 9.6

y = 6

put = y = 6 in equation (1)

36 + 6.25 = x²

42.25 = x²

⇒ x = 6.5

So, length of ladder is 6.5 m