NDA Mock Test: Mathematics - 10 - NDA MCQ

Given:

If the man moved 6 km/h faster, he would have taken 4 hours less. If he had moved 8 km/h slower, he would have taken 10 hours more.

Concept Used:

Distance = (Usual Speed × Increased speed)/(Diff. of speed) × Time

Calculation:

Let the original speed be S km/h.

[S(S + 6)/6] × 4 = [S(S - 8)/8] × 10

⇒ [(S + 6)/3] × 2 = [(S - 8)/4] × 5

⇒ 8S + 48 = 15S - 120

⇒ 7S = 168

S = 24 km/hr

Distance = [24(24 + 6)/6] × 4

= 480 km

Time = 480/24 = 20 hours.

∴ The answer is 20 hours.

Given:

A train starts at 3:00 pm from Mumbai and moves towards Ahmedabad at the speed of 20 km/hr.

Another train starts from Ahmedabad at 8 pm and moves towards Mumbai at the speed of 60 km/hr.

The distance between Mumbai and Ahmedabad is 600 km

Formula used:

Distance = speed × time

Concept used:

Relative speed for opposite direction = sum of the speed given

Calculation:

At 8 pm the man who started at 3 pm traveled 20 × 5 = 100 km

Now distance left 600 - 100 = 500 km

Realtive speed is 20 + 60 = 80 km/hr

The time will be ,

(500 /80) = 25/ 4 hr

So will meet after 6 hr 15 minutes from 8pm

So the exact time will be 8 pm + 6 hr 15 = 2:15 am

∴ The correct option is 4

Given:

The first scheme is a series of discounts of 10% and 20%

The second scheme is a series of discounts of 15% and 15%

Formula Used:

Effective discount = x + y - xy/100

where x and y are

Calculations:

Using the above formula,

Scheme (1)

Effective Discount = 10 + 20 - (10 × 20)/100

= 30 - 2

= 28%

Scheme (2)

Effective Discount = 15 + 15 - (15 × 15)/100

Effective Discount = 30 - 2.25 = 27.75%

Difference = 28 - 27.75 = 0.25%

The answer is first scheme, 0.25%.

Given:

5 men and 12 women can complete a work in 2 days

3 men and 4 women can complete it in 5 days.

Calculation:

According to question,

(5 men + 12 women) × 2 = (3 men + 4 women) × 5

⇒ 10 men + 24 women = 15 men + 20 women

⇒ 5 men = 4 women

⇒ men/women = 4/5

So, men : women = 4 : 5 i.e ratio of efficiency

Now, total work = (5 × 4 + 12 × 5) × 2

⇒ (20 + 60) × 2

⇒ 160 units

So, to complete 160 units in 8 days total men need = 160/(8 × 4) [Each man has an efficiency of 4 so we will multiply 4 with the time]

⇒ 5

∴ Required number of men is 5.

Given:

Percentage of various brands of LED TVs sold in the year 2020.

Calculations:

Total LED TVs sold = 1.5 lakh units.

Percentage of TVs sold by Bb brand = 23%

Number of TVs sold = 23% of 1.5 lakh units

⇒ 150000 × (23/100)

⇒ (1500 × 23) = 34500

∴ The answer is 34500.

Given:

The equation 2x2 + 5x + 5 = 0

Concept Used:

The quadratic equation Ax² + Bx + C = 0

If roots are imaginary, B² - 4ac < 0

Calculation:

Comparing of given equation

Now, A = 2, B = 5 & C = 5

⇒ (5)² - 4 × 2 × 5 < 0

⇒ 25 - 40 < 0

⇒ -15 < 0

∴ The roots are imaginary and distinct.

Concept:

• Modulus of function gives the magnitude and absolute value of a number and it is a non-negative value.
• The Square of any number is always positive.
• Also, the value inside the square root cannot be negative.

Calculation:

Given equation is

We know that modulus, square and square root are the positive terms, and sum of positive can be zero only if each term equals zero. Hence,

|x² - 1| = 0 ⇒ x = 1

(x - 1)² = 0 ⇒ x = 1

⇒ x² - x - 2x + 2 = 0

⇒ x(x - 1) - 2(x - 1) = 0

⇒ (x - 1)(x - 2) = 0

⇒ x = 1 and 2

Hence, x = 1 is the only solution which setisfy all above condition.

Given:

if x2 − 2x + 1 is a factor of 2x4 + x3 − 14x2 + 5k + 6.

Calculation:

if x2 − 2x + 1 is a factor of 2x4 + x3 − 14x2 + 5k + 6, then

Factors of x2 − 2x + 1 = (x - 1)² = (x - 1) (x - 1)

∴ x = 1 is the factor of 2x4 + x3 − 14x2 + 5k + 6.

Put x = 1 in this polynomial,

⇒ 2(1)⁴ + (1)³ - 14(1)² + 5k + 6 = 0

⇒ 2 + 1 - 14 + 5k + 6 = 0

⇒ 5k = 5

∴ k = 1

Given:

x2 + y2 + 1 = 2x

Formula used:

(a ± b)² = a² + b² ± 2ab

Calculation:

x2 + y2 + 1 = 2x,

⇒ x2 - 2x + 1 + y² = 0

⇒ (x - 1)² + y² = 0

In the above eq. the L.H.S. can only become zero when the base of terms i.e., (x – 1) and y becomes zero because for any other value the sum of their squares will always be a positive integer

Taking (x – 1) = 0 and y = 0

Therefore, x = 1 and y = 0

∴ x3 + y5 = 1 + 0 = 1.

Formula used:

(a + b)² = a² + 2ab + b²

(a - b)2 = a2 - 2ab + b2

Calculation:

Let,

y =

⇒ y =

⇒ y =

⇒ y =

⇒ y =

⇒ y =

We can see that, when 1 < x < 2,

Value of x - 1 is positive

Hence, option 3 is correct.

Given:

The product 101 × 102 × 103 × 197 × 198 × 199, find the last two digits.

Calculation:

Let's calculate the last two digits of each multiplication step by step, focusing only on the last two digits:

101 × 102, Last two digits = 01 × 02 = 02

Result × 103, Last two digits = 02 × 03 = 06

Result × 197, last two digits of 197 is 97.

So, 06 × 97. Last two digits = 6 × 97 = 582 = 82

82 × 198, Just use the last two digits of 198 is 98.

Last two digits = 82 × 98 = 8036 = 36

36 × 199, Again, only consider the last two digits of 199 is 99.

36 × 99 = 3564 = 64

∴ The last two digits of the entire product are 64.

Formula used:

Area of sector = πr² (θ/360º)

Circumference of the circle = 2πr

Calculation:

Let Radius of first, second & third circles be R₁, R₂ & R respectively.

It is given that,

⇒ R₁ = 0.17 m = 17 cm

⇒ R₂ = 9 cm

⇒ θ = 60°

Circumference of third circle = Circumference of first circle + Circumference of second circle

⇒ 2πR = 2πR₁ + 2πR₂

⇒ R = R₁ + R₂

⇒ R = 17 + 9 = 26 cm

Hence, radius of the third circle is 26 cm.

Area of sector = πR²(θ​​​​​​​/360º)

⇒ π(26)²(60​​​​​​​/360) = 112.67π cm² = 354.095 cm²

∴ The correct answer is 354.095 cm2.

Formula used:

Circumference of the circle = 2πr

Calculation:

Let Radius of first, second & third circles be R₁, R₂ & R respectively.

It is given that,

⇒ R₁ = 0.17 m = 17 cm

⇒ R₂ = 9 cm

Circumference of third circle = Circumference of first circle + Circumference of second circle

⇒ 2πR = 2πR₁ + 2πR₂

⇒ R = R₁ + R₂

⇒ R = 17 + 9 = 26 cm

∴ The correct answer is 26 cm.

Given:

EC = 15 m

DG = 5 m

Concept used:

Area of Trapezium = 1/2 × (a + b) × h,

Calculation:

GC = AC - AG = 25 - 12 = 13 m

⇒ Area = 1/2 × (15 + 5) × 13

⇒ 1/2 × 20 × 13

⇒ 130 m²

∴ The area of Trapezum DGCE is 130 m²

Given:

The area of the circular base of the tent house is 308 m^2, and its height is 7√2 m.

Concept Used:

The area of the circular base of the tent house = π × r²

The curved surface area of the conical tent = π × r × l

The slant height (L)² =

Where, r is radius & h is height.

Calculation:

According to question

π × r² = 308 m²

⇒ 22/7 × r² = 308 m²

⇒ r² = 308 × 7/22

⇒ r² = 98

⇒ r = 7√2 m

The slant height (L) =

⇒ L =

⇒ L=

⇒ L = 14 m

The curved surface area of the conical tent = π × 7√2 × 14

⇒ 22/7 × 7√2 × 14

⇒ 308√2 m

∴ The curved surface area of the conical tent is 308√2 m²

Calculation:

Let's solve the equation cos² θ - 2 + cos θ = 0 for the value of θ.

To make it easier, let's substitute cos θ with a variable, say x. The equation becomes:

x² - 2 + x = 0

Rearranging the equation:

x² + x - 2 = 0

Now, we can factorize the quadratic equation:

(x + 2)(x - 1) = 0

Setting each factor equal to zero:

x + 2 = 0 or x - 1 = 0

Solving for x:

x = -2 or x = 1

Since we substituted x with cos θ, we can replace x with cos θ in the solutions:

cos θ = -2 or cos θ = 1

However, the range of values for cosine is -1 ≤ cos θ ≤ 1. Therefore, cos θ cannot equal -2.

Hence, the value of θ that satisfies the equation cos² θ - 2 + cos θ = 0 is when cos θ equals 1.

Therefore, θ = 0° or θ = 360° (and any other angle measures that are equivalent to 0° or 360°).

∴ Option 4 is the correct answer.

Calculation:

33% of A is equal to 55% of B

⇒ A : B = 5 : 3.

Given

A person invests on simple interest = Rs. 7000

A person invests on simple interest = Rs. 5,000

Rate of Interest = R%

Rate of Interest = 5%

Time = 2 year

Time = 14 year

Formula used.

Simple interest = Principal × Time × Rate/100

Amount = Principal(1 - Rate/100)2

Calculation

The Rate of Interest for the first and second sum is R% and 5% p.a. respectively.

(7000 × R × 2) / 100 = ((5000) × 14 × 5) / 100

⇒ 140R = 3500

⇒ R = 3500/140

⇒ R = 25

The value of R (in percentage) is 25.

Given

Principal (P) = ₹20,000

Rate of interest (R) = 25%

Annual part payment = ₹2,500

Formula Used

Amount after n years = P(1 + R/100)ⁿ - Part payment using the structure of compound interest

Calculation

For the first year:

⇒ Amount after 1 year = 20,000 × (1 + 25/100)¹ - 2,500

⇒ A1 = 20,000 × 1.25 - 2,500

⇒ A1 = 25,000 - 2,500

⇒ A1 = 22,500

For the second year ⇒ P2 = 22,500

⇒ Amount after 2nd year = 22,500 × (1 + 25/100)¹ - 2,500

⇒ A2 = 22,500 × 1.25 - 2,500

⇒ A2 = 28,125 - 2,500

⇒ A2 = 25,625

The amount owed after two installments is ₹25,625.

Explanation -

We have,

+ +

= log_abc ab + log_abc bc + log_abc ca

= logabc (ab × bc × ca) = log_abc (abc)² = 2 logabc (abc) = 2

Concept -

We know that if g(x) is inverse of a bijection f(x), then

fog(x) = x ⇒ f(g(x)) = x

Explanation -

Clearly, f : [4, ∞) → [4, ∞) is a bijection.

So, it is invertible.

Let f(x) = y. Then,

5^{x(x - 4)} = y

⇒ x² - 4x = log₅ y

⇒ x² - 4x - log₅ y = 0

⇒

⇒ [∵ x ≥ 4]

Hence, .

We have,

⇒ g(f(x)) = 4x² - 10x + 4

⇒ (f(x))² + f(x) - 2 = 4x² - 10x + 4

⇒ (f(x))² + f(x) - (4x² - 10x + 6) = 0

⇒ f(x) =

⇒ f(x) =

⇒ f(x) = = 2x - 3, -2x + 2

Hence, f(x) = 2x - 3.

Given:

If P is subtracted from 8, 6, 2, and 9 then these numbers are in proportion.

Concept used:

If a, b, c and d are in proportion then

⇒ a/b = c/d

Mean proportion = √(a × b)

Calculation:

According to the question:

⇒ (8 - P)/(6 - P) = (2 - P)/(9 - P)

⇒ (8 - P) × (9 - P) = (2 - P) × (6 - P)

⇒ 72 - 8P - 9P + P² = 12 - 2P - 6P + P2

⇒ 17P - 8P = 72 - 12

⇒ 9P = 60

⇒ P = 20/3

Mean proportion = √{(3P - 6) × (9P - 4)}

Now, Putting the value of P in the equation:

⇒ √{(3 × (20/3) - 6) × (9 × (20/3) - 4)}

⇒ √{(20 - 6) × (60 - 4)}

⇒ √{14 × 56}

⇒ 14 × 2 = 28

∴ The correct answer is 28.

Given:

Number of red marbles Gagan has = 18

Number of blue marbles Gagan has = 12

Total number marbles Jane has = 20

Calculation:

Ratio of red and blue marbles Gagan has = 18/12

Ratio = 3/2 = 3 : 2

Jane has marbles in same ratio, then

Number of red marbles Jane has = 20 x (3/5) = 12

Number of blue marbles Jane has = 20 x (2/5) = 8

Difference between blue marbles both have = 12 - 8 = 4

∴ Gagan has 4 more blue marbles than Jane.

Concept used:

Efficiency = Total Work/ Total Time

Calculation:

Let the total capacity of the tank = 72 units [LCM of 12, 18 and 36]

According to the question

Efficiency of Pipe A = 72/12 = 6

Efficiency of Pipe B = 72/18 = 4

Efficiency of Pipe C = 72/36 = -2 (outlet pipe)

Combined efficiency of (A + B + C) = 6 + 4 + (-2) = 8

So, Time taken by (A + B + C) to fill the tank = 72/8 = 9 minutes

The answer is 9 minutes.

Given:

Distance = 30 km

Amit's time is 2 hours more than Sameer's time.

Amit's time with double speed is 1 hour less than Sameer's time.

Concept used:

Distance = Speed × Time

Calculation:

Let, Sameeer's speed = s km/hr

Sameer's required time = 30/s hr

According to the question,

Amit's required time = 30/s + 2 = (30 + 2s)s hr

So, Amit's speed = 30 ÷ (30 + 2s)/s = 30s/(30 + 2s) km/hr

30 ÷ (2 × 30s/(30 + 2s)) = 30/s - 1

⇒ 30 ÷ 60s/(30 + 2s) = (30 - s)/s

⇒ (30 + 2s)/2s = (30 - s)/s

⇒ s × (30 + 2s) = 2s × (30 - s)

⇒ 30s + 2s² = 60s - 2s2

⇒ 4s² = 30s

⇒ s = 7.5 km/hr

So, Amit's speed = (30 × 7.5)/(30 + 2 × 7.5) = 5 km/hr

∴ Amit's original speed is 5 km/hr.

Concept:

Regression coefficients b_yx =

Regression coefficients b_xy =

Calculation:

We know that b_yx =

⇒ b_yx = = -0.216

⇒ b_xy =

⇒ b_xy = = -0.129

∴ The regression coefficients byx and bxy are -0.216 and -0.129, respectively.

Concept:

The line of regression of y on x is y - y̅ = b_yx(x - x̅)

The line of regression of x on y is x - x̅ = b_xy(y - y̅)

Calculation:

The line of regression of y on x is y - y̅ = b_yx(x - x̅)

⇒ y - (29/5) = -0.216 (x - (24/5))

⇒ y = 6.836 - 0.216x

The line of regression of x on y is x - x̅ = b_xy(y - y̅)

⇒ x - (24/5) = -0.129 (y - (29/5))

⇒ x = 5.548 - 0.129y

∴ The two lines of regression are x = 5.548 - 0.129y and y = 6.836 - 0.216x.

Given:

a₁, a₂, a₃ ... be in AP such that

a₁ + a₅ + a₁₀ + a₁₅ + a₂₀ + a₂₅ + a₃₀ + a₃₄ = 300.

Concept:

If T₁, T₂, T₃ ......T_n are in AP then

T₁ + T_n = T₂ + T_n-1 = T₃ + T_n-2 = T₄ + T_n-4

Calculation:

We have

a₁ + a₅ - a₁₀ - a₁₅ - a₂₀ - a₂₅ + a₃₀ + a₃₄

⇒ (a₁ + a₃₄) + (a₅ + a₃₀) - (a₁₀ + a₂₅) - (a₁₅ + a₂₀)

Since, a₁ + a₃₄ = a₅ + a₃₀ = a₁₀ + a₂₅ = a₁₅ + a₂₀

∴ a₁ + a₅ - a₁₀ - a₁₅ - a₂₀ - a₂₅ + a₃₀ + a₃₄ = 0.

Calculation:

Average income is ,

⇒ (55 + 50 + 35 + 40 + 60) / 5

⇒ 240 / 5 = 48

Average expenditure is ,

⇒ (50 + 40 + 32 + 35 + 50) / 5

⇒ 207 / 5 = 41.4

So the difference is 48 - 41.4 = 6.6

∴ The correct option is 3.