Test: Elementary Mathematics - 8 - CDS MCQ

Given:

Mean of 10, 4, 5, 3, 2, 6, 7, 3 is m.

Mean of 2, 5, 3, 11, m, 10 is p.

Formula used :

Mean = Sum of all observation/Number of observation

Calculation:

According to question,

⇒ 10 + 4 + 5 + 3 + 2 + 6 + 7 + 3 = 8m

⇒ m= 5

Now,

⇒ 2 + 5 + 3 + 11+ m + 10 = 6p

⇒ p = 6

Now, (m + p) = 11

∴ The correct answer is 11.

Given:

AM = 4 cm and MC = 3 cm

∠AMC = 90°

Concept used:

C² = A² + B²

Calculation:

∠AMC = 90°

⇒ AC² = MC² + AM²

⇒ AC² = 3 + 4

⇒ AC² = 9 + 16

⇒ AC² = 25

⇒ AC = 5 cm

∴ The length of AC is 5 cm.

Given

sin α + cos α = ,

Formula used

sin² α + cos² α = 1

Calculation

sin α + cos α =

Squaring both sides,

sin2 α + cos2 α + 2sin α.cos α = (2/√3)²

1 + 2sin α.cos α = 4/3

2sin α.cos α = 4/3 - 1 = 1/3

sin α.cos α = 1/6

tan α + cot α = (sin α/cos α) + (cos α/sin α)

= (sin2 α + cos2 α )/(sin α.cos α)

= 1/(sin α.cos α) = 6

The answer is 6.

Given:

Radius of big ladoo, R = 810 cm

Radius of small ladoo, r = 90 cm

Formula used:

The volume of a sphere = (4/3)πR³

The surface area of a sphere = 4πR²

Calculation:

Volume of big ladoo = Volume of all small ladoos together

⇒ (4/3)πR³ = n × (4/3)πr³

⇒ n = (R/r)³ (Where n is the number of small ladoos)

Total surface area of all small ladoos = n × 4πr² = (R/r)³ × 4πr²

The ratio of total surface area of all small ladoos to the surface area of big ladoo

[(R/r)³ × 4πr²] : 4πR² = R/r = 810 : 90 = 9 : 1

∴ The correct answer is 9 : 1

Given:

r = 25 cm

R = 29 cm

Length of Common Chord = 40 cm

Calculation:

CE² = AC² – AE²

⇒ CE² = 25² – 20²

⇒ CE² = 625 – 400

⇒ CE² = 225

⇒ CE = √225

⇒ CE = 15 cm

ED² = AD² – AE²

⇒ ED² = 29² – 20²

⇒ ED² = 841 – 400

⇒ ED² = 441

⇒ ED = √441

⇒ ED = 21 cm

∴ The distance between the centres of circles i.e CD = (15 + 21) = 36 cm

The distance between the centres of circles is 36 cm.

Given:

Two zeroes of the polynomial x³ + 7x² - 2x - 14 are √2 and -√2.

Concept Used:

If α, β​​​​​​​, γ are the zeroes of the cubic polynomial p(x) = ax³ + bx² + cx + d

Then,

Solution:

We have,

A cubic polynomial as x³ + 7x² - 2x - 14 and two of its zeroes are √2 and -√2.

We know that,

If α​​​​​​​, β​​​​​​​, γ are the zeroes of the cubic polynomial p(x) = ax³ + bx² + cx + d

Then,

Here, a = 1, b = 7, c = - 2, d = - 14 and and

So,

⇒ γ = - 7 is the third zero for the cubic polynomial.

Option 2 is correct.

Shortcut Trick

A batsman scored X runs in his nth match. The average run scored by the batsman in (n - 1) matches is Y. If the average run scored by a batsman is increased by Z runs,

The new average after his nth match = X - Z(n - 1)

According to the question,

X = 135

Z = 5

(n - 1) = 11

Therefore,

The new average after his nth match = 135 - Z(11) = 80

∴ The required answer is 80.

Alternate Method

Given:

A batsman scored 135 runs in his 12th match.

If the average run scored by the batsman in the last 11 matches is x,

Calculation:

The average run in 11 matches = x

The total run scored in 11 matches = 11x

The run scored in 12th match = 135

The total run in all 12 matches = 11x + 135

Now,

The new average = x + 5

Therefore,

⇒ = x + 5

⇒ 11x + 135 = 12x + 60

⇒ x = 75

Therefore,

The new average = 75 + 5 = 80

∴ The required answer is 80.

Given:

Speed of the speedboat in still water = 20 km/h

Distance downstream = 35 km

Total time for the round trip = 8 hours

Calculation:

Let the speed of the stream be (v) km/h.

Downstream Speed= (20 + v) km/h

Upstream Speed = (20 - v) km/h

Time for downstream = (35/(20 + v))

Time for upstream = (35/(20 - v))

Total Time = Time downstream + Time upstream

⇒ (35 / (20 + v) + 35 / (20 - v) = 8) hours

⇒ 35 / (20 + v) + 35 / (20 - v) = 8

⇒ 35(20 - v) + 35(20 + v) = 8(20² - v²)

⇒ 700 - 35v + 700 + 35v = 8(400 - v2)

⇒ 1400 = 3200 - 8v2

⇒ 8v2 = 3200 - 1400

⇒8v2 = 1800

⇒ v2 = 225

⇒ v = 15

Therefore, the speed of the stream is 15 km/h.

Calculation

Let PD be x cm, then QD = (18 – x) cm

PD = PF, QD = QE, RF = RE [tangents]

⇒ QE = (18 – x) cm

⇒ RE = 13 – (18 – x) = x – 5

⇒ RF = (x – 5)

⇒ PF = 15 – (x - 5) = 20 – x

⇒ 20 – x = x

⇒ 2x = 20

⇒ x = 20/2

⇒ x = 10 cm

⇒ PD = 10

The answer is 10.

Given:

PA = x² + x - 8

PB = x² - x

OP = 13 cm

Concept used:

The two tangents drawn from a common external point to a circle are of equal length.

Each tangent to a circle is perpendicular to the radius at the point of tangency.

Circumference of the circle = 2πr

Area of the equilateral triangle = (√3/4)a²

Perimeter of the equilateral triangle = 3a

Where,

r is the radius of the circle and a is the side of the equilateral triangle.

Calculation:

As we know, PA = PB

⇒ x² + x - 8 = x² - x

⇒ 2x = 8

⇒ x = 4

​⇒ PA = 4² + 4 - 8 = 12

​⇒ PA = 12 cm

Using Pythagoras theorem,

⇒ PA² + OA² = OP²

⇒ 12² + OA² = 13²

⇒ OA² = 13² - 12²

⇒ OA = √(13² - 12²)

⇒ OA = 5 cm

Radius of the new circle = (OA + 16)

⇒ (5 + 16) = 21 cm

According to the question,

Perimeter of equilateral triangle = circumference of the circle

⇒ 3a = 2 × (22/7) × 21

⇒ a = 44 cm

Area of the equilateral triangle is = (√3/4)× 44²

⇒ 484√3 cm²

∴ The area of the equilateral triangle is 484√3 cm².

Shortcut Trick

13 ∶ 21 ∶ 10 ∶ 25

13x + 21x + 10x + 25x = 69 × 4

69x = 69×4

x = 4

The difference (25 - 10 = 15) = 15x = 15×4 = 60

Alternate Method

Given:

The age of 4 persons = 13 ∶ 21 ∶ 10 ∶ 25

Average age = 69

Concept:

The average age of four persons is the sum of their ages divided by 4.

Formula used:

Average = (Sum of all numbers)/(Sum of all numbers)

Calculation:

Let's denote the ages of the four persons as 13x, 21x, 10x, and 25x, respectively, where x is a common factor.

According to the given information:

Sum of ages = 13x + 21x + 10x + 25x = 69 × 4

We can solve for x

⇒ 13x + 21x + 10x + 25x = 69×4

⇒ 69x = 69×4

⇒ x = 4

So, the ages of the four persons are

⇒ 13x = 13×4 = 52

⇒ 21x = 21×4 =84

⇒ 10x = 10×4 = 40 = the age of the youngest

⇒ 25x = 25×4 = 100 = the age of the oldest

The difference the age = 100 - 40 = 60

∴ The difference between the age of the oldest and the youngest person is 60 years

Given :

The cost increases by 60% whereas the selling price remains the same

Profit% = 540%

Formula Used :

Profit% = [(S.P. - C.P.)/C.P.] × 100

C.P. = cost price

S.P. = selling price

Calculation :

Let CP be 100.

SP = 100 × 640/100 = 640

New CP = 100 × 160/100 = 160

New profit = 640 - 160 = 480

Profit:SP = 480:640 = 3:4

The answer is 3:4.

Given:

9a3 + 8b3 - c3 + a3 - 4b3 + 3

Concept used:

A coefficient refers to a number or quantity placed with a variable.

Calculation:

Here, the co-efficient of c³ is (-1).

∴ The co-efficient of c3 is (-1).

Given:

AM is perpendicular to BC and AM is the bisector of ∠A.

∠A = 60°.

Concept used:

The sum of interior angles of a triangle = 180°

Exterior angle = 180° - interior angle

Calculation:

⇒ ∠BAM = (60º/2) = 30°

⇒ ∠BAM + ∠AMB + ∠B = 180°

⇒ 30° + 90° + ∠B = 180°

⇒ ∠B = 180° - 120°

⇒ ∠B = 60°

Exterior ∠B = 180° - 60° = 120°

∴ The measure of exterior ∠B is 120°.

Concept:

Rules for Operations on Inequalities:

• Adding the same number to each side of an inequality does not change the direction of the inequality symbol.
• Subtracting the same number from each side of an inequality does not change the direction of the inequality symbol.
• Multiplying each side of an inequality by a positive number does not change the direction of the inequality symbol.
• Multiplying each side of an inequality by a negative number reverses the direction of the inequality symbol.
• Dividing each side of an inequality by a positive number does not change the direction of the inequality symbol.
• Dividing each side of an inequality by a negative number reverses the direction of the inequality symbol.

​Solution:

Given that,

⇒

⇒

⇒

Cross-multiplying, we have

20(2x - 1) ≥ 3(19x - 18)

⇒ 40x - 20 ≥ 57x - 54

⇒ 40x - 57x ≥ 20 - 54

⇒ -17x ≥ -34

⇒

⇒ x ≤ 2

∴ The solution set =

Hence, The positive integral solutions are 1 and 2.

So, the number of positive integral solutions of the given inequality is 2.

∴ The correct option is (1)

Given:

Two persons Ram and Shyam can dig a pit in 25 days and 30 days respectively.

Mohan can fill that pit in 75 days

The total time taken to dig the pit from the beginning is 15 days,

Calculation:

Let total work be LCM of 25, 30, and 75 days = 150 units

One day's work of Ram, Shyam, and Mohan is 6 units, 5 units, and 2 units.

But the nature of the work of Mohan is opposite to that of Ram and Shyam.

Let Mohan work for ‘x’ days and the remaining days in which only Ram and shyam work is (15 – x) days.

According to the question

⇒ x × (6 + 5 - 2) + (15 - x) × (6 + 5) = 150

⇒ 9x + (15 - x) × 11 = 150

⇒ 9x + 165 - 11x = 150

⇒ 2x = 15

⇒ x = 7.5

∴ The correct answer is 7.5 days

Alternate Method

Total work will be LCM of (25, 30, and 75) = 150 units

Efficiency of Ram = 150/25 = 6

Efficiency of Shyam = 150/30 = 5

Efficiency of Mohan = 150/75 = 2

Since Ram and Shyam do the work for 15 days.

Total work done by Ram and Shyam = 15 × (6 + 5) = 165 units

the But total work to be done is = 150

The negative work done by Mohan = 165 - 150 = 15 units

Time taken by Mohan to complete the work = 15/Efficiency

⇒ 15/2 = 7.5 days

∴ The correct answer is 7.5 days.

Formula Used :

For a quadratic polynomial ax² + bx + c

Sum of zeroes = - b/a, Product of zeroes = c/a

Calculation :

Using formula,

Sum of zeroes α + β = - (- 6/2) = 3 ------(1)

Product of zeroes α × β = - 15/2 ------(2)

Now, (1/α) + (1/β) = (α + β)/(α × β ) -------(3)

From equation(1), (2) & (3)

(1/α) + (1/β) = 3/(- 15/2) = - 2/5

∴ The value of (1/α) + (1/β) = - 2/5.

The correct answer is 50 kg
Key Points

The average weight of a class of 15 students is 42. ⇒ Total weight of these 42 students is (42 × 15) = 630 kg
If the teacher is included, the number of persons concerned increases to 16. And, the average weight also increases to 16 kg (as given).
⇒ The sum of the weight of 16 people is (43 × 16) = 688 kg
Hence, the required weight of the teacher is (688 - 630) kg = 50 kg

Concept:

1). Geometric Progression

The general form of Geometric Progression is:

a, ar, ar2, ar3, ar4,…, arn-1

where, a = First term, r = common ratio, arn-1 = nth term.

2). The sum of n terms of GP is given by:

Sn = a + ar + ar2 + ar3 +…+ arn-1

Sn =

Calculation:

Let a = first term and r = common ratio.

ar + ar2 + ar3 = 3 and ar5 + ar6 + ar7 = 243

⇒

⇒

⇒ r = 3

Putting the value of r in ar + ar2 + ar3 = 3

⇒ a(3 + 9 + 27) = 3

⇒ a = 3/39 = 1/13

Sum of the first 50 terms = S₅₀ =

⇒ S50 =

⇒ S50 =

∴ The sum of the first 50 terms is

Given:

f(x) = 3x⁴ - 9x³ + x² + 15x + k

Solution:

k + 10 = 0

k = -10

Hence, the correct option is 1.

Given:

The ratio of boys and girls in a college is 5 : 3

24% of boys and 32% of girls are not placed in campus placements.

Calculation:

Let the total number of students be 800.

The number of boys = 800 × 5/8 = 500

The number of girls = 800 × 3/8 = 300

According to the question,

24% of boys and 32% of girls are not placed in campus placements.

The percentage of boys who got jobs in campus placements = 100% - 24% = 76%

⇒ 76% of 500 = 380

The percentage of girls who got jobs in campus placements = 100% - 32% = 68%

⇒ 68% of 300 = 204

Total placements = 380 + 204 = 584

∴ Percentage of placements = (584/800) × 100 = 73%

∴ The percentage of those students who got jobs in campus placements is 73%.

Given:

Reena invested a total of Rs. 25,000.

Part of the investment is at 10% per year.

The other part of the investment is at 12% per year.

The total income from both investments for one year is Rs. 2800.

Formula Used:

Simple Interest = Principal × Rate × Time / 100

Calculation:

Let the amount invested at 10% be Rs. x.

Then, the amount invested at 12% will be Rs. (25000 - x).

Income from the amount invested at 10% = x × 10 × 1 / 100 = 0.1x

Income from the amount invested at 12% = (25000 - x) × 12 × 1 / 100 = 0.12 × (25000 - x)

According to the problem, the total income is Rs. 2800.

So,

0.1x + 0.12(25000 - x) = 2800

Solving the equation:

⇒ 0.1x + 3000 - 0.12x = 2800

⇒ 3000 - 2800 = 0.12x - 0.1x

⇒ 200 = 0.02x

⇒ x = 200 / 0.02

⇒ x = 10000

So, the amount invested at 10% is Rs. 10,000.

The amount invested at 12% is 25000 - 10000 = Rs. 15,000.

The correct answer is option 1: Rs. 10,000 at 10% and Rs. 15,000 at 12%

Given:

The thief started cycling at 7:30 pm
The policeman started half an hour later at 8:00 pm
The thief's speed is 10 km/h.
The policeman's speed is 12 km/h.

Formula used:

Distance = Speed × Time

Calculation:

Let t be the time it takes for the policeman to catch the thief.

In the half-hour interval between the thief's departure and the policeman's departure

The thief travels, Distance = Speed × Time

D = 10 × (1/2) km = 5km

So, when the policeman starts at 8:00 P.M., the thief is already 5 km away

Now,

Distance covered by the policeman = Relative speed × t

⇒ Relative speed = (12 - 10) = 2 km/h

Now,

⇒ 2km/h × t = 5 km

⇒ t = (5/2) = 2.5hours 
Since the policeman started at 8:00 P.M., he will catch the thief 2.5 hours later:

8:00 pm + 2:30 hours = 10:30 pm

∴ The policeman will catch the thief at 10:30 pm

Cost price of the bottle = 4500 Rs.

Marked price of the bottle = 4500 × [(100 + x)/100] Rs.

It is sold after two successive discount of Rs. 140 and 10% respectively for a profit of Rs. 720.

Thus,

SP = CP + Profit = MP - (Successive Discount of 140 & 10%)

{[4500 × [(100 + x)/100]] - 140} × 90% = 4500 + 720

{[4500 × [(100 + x)/100]] - 140} × 0.9 = 5220

[4500 × [(100 + x)/100] - 140 = 5800

[4500 × [(100 + x)/100] - 140 = 5940

[(100 + x)/100] = 5940/4500

⇒ 100 + x = 132

⇒ x = 32%

Cost price of the jeans = 80% × 4500 = 3600 Rs.

The selling price of the jeans = 4680 Rs.

Profit% = [(4680 - 3600)/3600] × 100 = 30%

ATQ, x - k = 30

⇒ k = 32 - 30

⇒ k = 2

Hence, the correct answer is 2.

Given:

M is the midpoint of BC

Calculation:

⇒ M divides BC into two equal parts.

⇒ BM = MC

⇒ BC = BM + MC

⇒ BC = k + k

⇒ BC = 2k

∴ The length of BC is 2k cm.

Concept:

Modulus: The absolute value or modulus of a real number x, denoted |x|, is the non-negative value of x without regard to its sign.

• Namely, |x| = x if x is positive, and |x| = −x if x is negative, and |0| = 0.
• For example, the absolute value of 3 is 3, and the absolute value of −3 is also 3
• Symbolically, modulus of x is given by,

​Calculations:

Given, |2x + 3| > 5

2x + 3 > 5 and - (2x + 3) > 5

First, consider 2x + 3 > 5

subtract 3 on both sides of inequality

⇒ 2x + 3 - 3 > 5 - 3

⇒ 2x > 2

⇒ x > 1

⇒ x ∈ (1, ∞) ......(1)

Next, consider - (2x + 3) > 5

⇒ - 2x - 3 > 5

add 3 on both sides of inequality

⇒ - 2x - 3 + 3 > 5 + 3

⇒ - 2x > 8

⇒ - x > 4

⇒ x < - 4

⇒ x ∈ (- ∞, - 4) ......(2)

From (1) and (2), we have

x ∈ (-∞, -4) ∪ (1, ∞) is the solution set of |2x + 3| > 5

Hence, the correct answer is option 3.

Given:

The ratio of the number of coins = 3:5:7

Value of coins = Rs. 1, Rs. 5, and Rs. 10 respectively.

Total value of coins = Rs. 392

Formula Used:

Total value = (Number of Rs. 1 coins × 1) + (Number of Rs. 5 coins × 5) + (Number of Rs. 10 coins × 10)

Calculation:

Let the number of Rs. 1, Rs. 5, and Rs. 10 coins be 3x, 5x, and 7x respectively.

⇒ Total value = (3x × 1) + (5x × 5) + (7x × 10)

⇒ 3x + 25x + 70x = 392

⇒ 98x = 392

⇒ x = 392 / 98

⇒ x = 4

⇒ Number of Rs. 10 coins = 7x = 7 × 4

⇒ 28

∴ The total number of Rs. 10 coins is 28.

Concept:

In the quadratic equation ax² + bx + c = 0

Sum of root (α + β) = -b/a

Product of root (αβ) = c/a

Formula used:

(x + y)² = x² + y² + 2xy

(x - y)2 = x2 + y2 - 2xy

Explanation:

We have x2 + ax + b = 0

By using the above concept

α + β = - a -----(1)

α.β = b -----(2)

To find α, we need to consider the given statements.

Statement 1:

α + β = 0 or β = - α

Also, α2 + β2 = 2

⇒ α2 + (-α)2 = 2

⇒ 2α² = 2

⇒ α = ± 1

Statement 2: αβ² = -1, a = 0

From equation (1)

α + β = - a

α + β = 0 [∵ a = 0]

⇒ β = - α

Now

⇒ αβ² = -1

⇒ α (-α)² = -1

⇒ α³ = -1

⇒ α = -1

∴ The correct answer is option (3).

Given:

The average weight of a class of students = 67.5 kg

Formula used:

[( Weight of teacher - Average weight of the class ) / Weight of the teacher] × 100

Calculation:

Let, The weight of the class teacher = Y

According to the questions,

The weight of the class teacher = 25% more than the average weight of the class.

The weight of the class teacher = 67.5 + (25/100) × 67.5

The weight of the class teacher = 67.5 + 16.875

The weight of the class teacher = 84.375 kg

Now,

The average weight of the class is less than the class teacher by x%.

= [( Weight of teacher - Average weight of the class ) / Weight of the teacher] × 100

= [(84.375 - 67.5) / 84.375] × 100

= [16.875 / 84.375] × 100

= 20%

∴ The value of x is 20%.

Given:

The ratio of the length, width, and height of a cuboid = 6:3:2.

Total surface area of the cuboid = 1800 cm².

Formula Used:

Surface Area of a Cuboid = 2(lw + lh + wh)

Volume of a cuboid = l × w × h

Calculation:

Let the length = 6x, width = 3x, and height = 2x.

⇒ Surface area = 2[(6x × 3x) + (6x × 2x) + (3x × 2x)] = 1800

⇒ 2[18x² + 12x² + 6x²] = 1800

⇒ 2[36x²] = 1800

⇒ 72x² = 1800

⇒ x² = 1800 / 72

⇒ x² = 25

⇒ x = 5

⇒ Volume = l × w × h = (6x) × (3x) × (2x)

⇒ Volume = 6 × 3 × 2 × x³

⇒ Volume = 36 × 5³

⇒ Volume = 36 × 125

⇒ 4500 cm³

∴ The volume of the cuboid is 4500 cm³.