Test: Elementary Mathematics - 5 - CDS MCQ

We have,

log_ax x, log_bx x, log_cx x in H.P.

⇒ log_x ax, log_x, bx, log_x cx are in A.P.

⇒ log_x a + 1, log_x b + 1, log_x c + 1 are in A.P.

⇒ log, a, log, b, logx c are in A.P.

⇒ a, b, c are in G.P.

Hence Option (2) is Correct.

Given:

x2 - 15x + 1 = 0

Concept used:

if x + 1/x = P

then x² + 1/x² = P² - 2

Calculation:

x2 - 15x + 1 = 0

Divide by x on both sides

⇒ x - 15 + 1/x = 0

⇒ x + 1/x = 15

Now,

⇒ x² + 1/x² = 15² - 2

⇒ x2 + 1/x2 = 223

Now putting the value in the given expression

⇒ x4 - 223x2 + 6

⇒ x4 - (x2 + 1/x2) × x2 + 6

⇒ x4 - x⁴ - 1 + 6

⇒ 5

∴ The correct answer is 5.

Given:

k(21x² + 24) + rx + (14x² − 9) = 0 ----- (eq.1)

k(7x² + 8) + px + (2x² − 3) = 0 ------ (eq. 2)

Calculation:

Since the roots of the equation are common, the coefficients are in proportion.

The eq.1 can be written as, (21k +14)x² + rx + 24k - 9 = 0

The eq.2 can be written as, (7k + 2)x² + px + 8k - 3 = 0

Therefore,

⇒ = =

⇒ r/p =

⇒ r/p =

⇒ r/p = 3/1

⇒ p/r = 1/3

Hence, the correct answer is option 1.

Given:

cosecA + cotA = a√b

Concept:

We'll use trigonometric identities to manipulate the given equation and express it in terms of sinA and cosA

Formula Used:

cosecA = 1/sinA,

cotA = cosA/sinA

sin²A + cos²A = 1

Calculation :

cosecA + cotA = a√b

1/sinA + cosA/sinA = a√b

(1 + cosA)/sinA = a√b

Now, we'll square both sides of the equation

{(1 + cosA)/sinA}² = (a√b)²

__________Eqn 1

Subtracting -1 in both sides, we gets

= ________Eqn 2

Adding +1in both sides to eqn1

= _______ Eqn3

Now, dividing eqn2 and eqn3

cosA =

So, the answer is cosA

Calculation:

Total Toffees in 8 packets is 8 × 36 = 288

1/4 of toffees distributed among his friends = 288 × 1/4 = 72

3/8 of toffees distributed among his teachers = 288 × 3/8 = 108

1/3 of toffees she saved for his brother = 288 × 1/3 = 96

Toffees left with her = total toffees - distributed toffees = 288 - (72 + 108 + 96)

Toffees left with her= 288 - 276 = 12

Mistake Points

Here, distribution is happening on total number of toffees and not on the remaining toffees

⇒ x = (633)²⁴ – (277)³⁸+ (266)⁵⁴

⇒ x = 3²⁴ – 7³⁸ + 6⁵⁴

⇒ x = 3⁴ – 7² + 6²

⇒ x = 1 – 9 + 6

⇒ x = 7 – 9

Or x = 17 – 9

⇒ x = 8

Key Points

Unit digit of a number is digits in the one's place of the number.

It is rightmost of the number. like 523417, 7 is the unit digit.

Concept use:

The sum of the first n natural numbers can be found using the formula:

Sum = n × (n + 1) / 2

Where: n is the number of terms (in this case, 100)

Calculation:

Sum = 100 × (100 + 1) / 2
Sum = 100 × 101 / 2
Sum = 5050

Next,

we need to find the remainder when 5050 is divided by 77.

Remainder = 5050 / 77 × x where, x is the quotient
Remainder = 5050 - (77 × (5050 // 77))
Remainder = 5050 - (77 × 65)
Remainder = 5050 - 5005
Remainder = 45

Hence, the remainder left after dividing the sum of the first 100 natural numbers by 77 is 45.

Given:

PA = x² + 3x -168

PB = x² - 14x + 36

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.

Calculation:

As we know, PA = PB

⇒ x² + 3x -168 = x² - 14x + 36

⇒ 17x = 204

⇒ x = 12

⇒ PA = 12² + (3 × 12) -168

​⇒ PA = 12 cm

Using Pythagoras theorem,

⇒ PA² + OA² = OP²

⇒ 12² + OA² = 13²

⇒ OA² = 13² - 12²

⇒ OA = √(13² - 12²) = 5 cm

∴ The radius of the circle is 5 cm.

Given:

In Δ ABC , PQ is parallel to the side BC . AP = QC , Ab = 16cm and AQ = 4cm

Concept used:

In Δ ABC , PQ parallel to the side BC

So AP / PB = AQ /QC

Calculation:

Let AP = QC = x , PB = 16 - x

As per the question,

⇒ x² + 4x - 64 = 0

⇒ x =

⇒ x =

⇒ x =

⇒ x =

⇒ x =

⇒ x =

∴ The correct option is 3.

We have,

xyz + 1 = log_2a a × log_3a 2a × log_3a 4a + 1

⇒ xyz + 1 = log_4a a + 1

⇒ xyz + 1 = log_4a a + log_4a 4a = log_4a (2a)²

⇒ xyz + 1 = log^4a (2a)² = 2 log_4a 2a = 2 log_3a 2a × log_4a 3a

⇒ xyz + 1 = 2yz

Hence Option (1) is Correct.

Key Points

If 2 is added to each observation, the mean will also increase by 2. Therefore, the new mean will be 45 + 2 = 47.

• The standard deviation is a measure of how spread out the data is. Adding 2 to each observation will not change how spread out the data is. Therefore, the standard deviation will remain the same, which is 10.
• Therefore, the new mean and standard deviation will be 47 and 10, respectively.

The other options are not correct:

• 47, 12: This option is incorrect because the standard deviation will not change when 2 is added to each observation.
• 47, 14: This option is incorrect because the standard deviation will not increase by 4 when 2 is added to each observation.
• 46, 12: This option is incorrect because the mean will increase by 2 when 2 is added to each observation.

Therefore, the correct answer is (3) 47, 10.

Given:

Average marks of 21 papers = 50.

Average marks of the first 11 papers = 52.

Average marks of the last 11 papers = 45.

Formula Used:

Total Marks = Average × Number of Papers

Calculation:

Total marks of 21 papers ⇒ 50 × 21

Total marks of the first 11 papers ⇒ 52 × 11

Total marks of the last 11 papers ⇒ 45 × 11

Let the marks of the 11th paper be x.

The 11th paper is counted in both the first 11 and the last 11 papers.

So, the sum of the first 11 papers and the last 11 papers minus the 11th paper's marks should equal the total marks of the 21 papers.

(52 × 11) + (45 × 11) - x = 50 × 21

572 + 495 - x = 1050

1067 - x = 1050

x = 1067 - 1050

x = 17

The marks of the 11th paper is 17.

Given

x² - 5√ 5x + 1 = 0

Formula used

(a - b)² = (a + b)² - 4ab

(a - b)³ = a³ - b³ - 3ab(a - b)

Calculation

If we simplify the given equation,x2 - 5√ 5x + 1 = 0, by dividing it by x.

⇒ x + 1/x = 5√ 5

Applying the above-mentioned formula:

⇒ (x - 1/x)2 = (x + 1/x)2 - 4x × 1/x

⇒ (x - 1/x)2 = 125 - 4

⇒ (x - 1/x)2 = 121

⇒ x - 1/x = 11

⇒ x3 - 1/x³ = (x - 1/x)³ + 3x × 1/x(x + 1/x)

⇒ x3 - 1/x3 = (11)3 + 3(11)

⇒ x3 - 1/x3 = 1331 + 33

⇒ x3 - 1/x3 = 1364

The answer is 1364.

Given:

Salary spent on food = 30% of the income

Salary spent on household expenses = 50% of the remaining

Savings = Rs. 10,500

Calculation:

Let the total income of Ram be Rs. 100x

Salary spent on food = 30% of 100x = 30x

Remaining amount = (100x – 30x) = 70x

Salary spent on household expenses = 50% of 70x

⇒ 35x

Savings = 100x – (30x + 35x)

⇒ 35x

Now, 35x = Rs. 10,500

⇒ 100x = Rs. (10,500/35x) × 100x

⇒ Rs. 30,000

Now, if total salary of Shyam be 100%

Then, 75% = Rs. 30,000

⇒ 100% = Rs. (30,000/75) × 100

⇒ Rs. 40,000

∴ Salary of Shyam is Rs. 40,000

Given:

tanθ + cotθ = 2

θ is an acute angle (θ < 90^° )

Concept used:

At θ = 45^°, tanθ = cotθ = 1

Similarly, tanⁿθ = cotⁿθ = 1

Calculation:

We have

2 tan²⁵ θ + 3 cot²⁰ θ + 5 tan³⁰ θ cot¹⁵ θ

⇒ (2 × 1) + (3 × 1) + (5 × 1 × 1) = 10

∴ The correct answer is 10.

Given:

A, B, and C can individually complete a task in 12, 15, and 20 days respectively

Concept used:

Efficiency = Total work / Time taken

Calculation:

Let the total work be LCM of (12, 15 and 20) = 60

⇒ Efficiency of A = 60/12 = 5

⇒ Efficiency of B = 60/15 = 4

⇒ Efficiency of C = 60/20 = 3

Time taken to complete the work by A, B and C together is given,

⇒ Total work / (Sum of efficiencies of A, B and C)

⇒ 60/(5 + 4 + 3) = 60/12

⇒ 5 days

∴ The A, B and C working together complete the work in 5 days.

Given:

The proportion of milk and water in a 105 liters mixture is 3:4. This ratio is changed to 3:5.

Calculation:

Initial ratio of milk to water = 3:4

Total mixture = 105 liters

⇒ Initial quantity of milk = (3/7) × 105

⇒ Initial quantity of milk = 45 liters

⇒ Initial quantity of water = (4/7) × 105

⇒ Initial quantity of water = 60 liters

The new ratio of milk to water = 3:5

The total quantity of milk remains the same = 45 liters

⇒ Let the new total quantity of the mixture be X liters

⇒ (3/8) × X = 45

⇒ X = (45 × 8) / 3

⇒ X = 120 liters

New quantity of water = Total mixture - Quantity of milk

⇒ New quantity of water = 120 - 45

⇒ New quantity of water = 75 liters

Excess water added = New quantity of water - Initial quantity of water

⇒ Excess water added = 75 - 60

⇒ Excess water added = 15 liters

∴ The amount of excess water mixed in the mixture is 15 liters.

Given:

Ratio of different types of old coins = 3 : 5 : 7.

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

Total value of coins = Rs. 490.

Formula Used:

Total value = (Number of coins of type 1 × Value of type 1) + (Number of coins of type 2 × Value of type 2) + (Number of coins of type 3 × Value of type 3)

Calculation:

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

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

⇒ Total value = 3x + 25x + 70x

⇒ Total value = 98x

⇒ 98x = 490

⇒ x = 490 / 98

⇒ x = 5

Number of old coins of 10 rupees = 7x = 7 × 5 = 35

The number of old coins of 10 rupees is 35.

The correct answer is Option 1 i.e. Both 362 and 1000

Explanation-
In the question, look for the unit digit number and the power.
For example, the powers of 9 operate as follows:
9¹ = 9,
9² = 81,
9³ = 729, and so on.
Hence, the power cycle of 9 also contains only 2 numbers 9 & 1, which appear in case of odd and even powers respectively.

While that for 1 raised to any power will result in unit place being 1. When 1 is subtracted with 1 it results in a 0 at unit place, making the number divisible by both 362 and 1000.

Calculation:

Given x² - 5x - 14 > 0

⇒ (x + 2)(x - 7) > 0

⇒ x < - 2 or x > 7

x ∈ (- ∞, - 2) ∪ ( 7, ∞ )

x lies outside [ - 2, 7]

Therefore, α = - 2 and β = 7

⇒ α/β = (-2)/7

The correct answer is option (1).

The speed of a boat in still water is 5km/hr and the speed of the stream is 1 km/hr

Speed while going with the stream = speed in still water + speed in stream

= (5 + 1)

= 6 kmph

Speed while going against the stream = speed in still water - speed in stream

= (5 - 1)

= 4 kmph

Time = Distance / Speed

As distance is given = 24 km.

Time taken to cover the distance while going with the stream = (24/6)

= 4 hours.

Time taken to cover the distance while going against the stream = (24/4)

= 6 hours.

Total time taken = (4 + 6)

= 10 hrs.

Therefore, the correct answer is "10 hr".

Concept Used:

Change in % = (New value - old value)/old value × 100

Calculation:

Let the number be 100

Now, the number is increased by 20%, 25% and 30%

⇒ 100 × 120/100 × 125/100 × 130/100 = 195

Change in % = (New value - old value)/old value × 100

⇒ (195 - 100)/100 = 95%

∴ The single equivalent increase if the number is successively increased by 20%, 25% and 30% is 95%.

Calculation:

It is evident from the Venn-diagram that

(A - B) U (B - A) = (A U B) - (A ∩ B)

∴ ((A - B) U (B - A)) U (A ∩ B)

= ((A U B) - (A ∩ B)) U (A ∩ B)

= A U B

Correct answer is option 1.

Total surface area of the solid sphere = 4πR2

⇒ 4 × 3.15 × R2 = 315

⇒ R2 = 25

⇒ R = 5

Volume of the solid sphere = (4/3) × πR3

Volume of the solid sphere = (4/3) × 3.14 × (5)3 = 523.33 ≈ 525 cm3

Edge length of the cubical tank = 10 cm

Total Volume of the cubical tank = (10)3 = 1000 cm3

Current volume of the cubical tank = m% × 1000 cm3

After dropping the sphere into the tank, the height of the liquid in the tank increases by 175%.

Therefore, (m% × 1000 + 525)/100 = m% × (1000/100) × (11/4)

⇒ m% × 4000 + 2100 = m% × 11000

⇒ 7000 × m% = 2100

⇒ m% = 30%

Given:

X = {8n - 7n - 1 : n ∈ N}

Y = {49 (n - 1) : n ∈ N}

Formula used:

(1 +x)ⁿ =

Calculations:

We have,

8n - 7n - 1

⇒ (1 + 7)ⁿ - 7n - 1

⇒ - (7n + 1)

⇒ 1 + 7n + 7² - (7n + 1)

⇒ 7²

∴ X contains some multiples of 49.

Now,

49 (n - 1)

Clearly, Y contains all multiples of 49 including 0.

Thus, X ⊂ Y.

∴ The correct answer is option (1).

Given:

A = {(x, y) : x2 + y2 ≤ 1, x, y ∈ R)

B = {(x, y): x2 + y2 ≤ 4; x, y ∈ R)

Calculation:

Clearly, A is the set of all points lying inside or on the circle with centre at the origin and radius 1 and B is the set of all points lying inside or on the circle with centre at the origin and radius 2 units. Clearly, A ⊂ B.

Therefore, A - B = ϕ and B - A ≠ ϕ

∴ Correct answer is option 3.

Given:

The numerator and denominator of a fraction are in the ratio 2 ∶ 3

Calculation:

Let the proportionality constant is x

According to the question, The fraction is 2x/3x

⇒

⇒

⇒ 18x - 54 = 12x

⇒ 6x = 54

⇒ x = 9

Numerator = 2x = 2 × 9 = 18

∴ The numerator of the real fraction is 18.

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.

​Calculation:
Given:

First by solving the inequality: we get

⇒ 4 ≤ 3(x + 1)

⇒ 4 ≤ 3x + 3

⇒ 4 - 3 ≤ 3x

⇒ 1 ≤ 3x

⇒ x ≥ (1/3) .....(1)

Similarly, by solving the inequality:

⇒ 3(x + 1) ≤ 6

⇒ 3x + 3 ≤ 6

⇒ 3x ≤ 6 - 3

⇒ 3x ≤ 3

⇒ x ≤ 1 .....(2)

From equations (1) and (2) we can say that (1/3) ≤ x ≤1 

So, out of the given options, the possible value which x can take is 2/3.

∴ The correct option is (3).

Concept used

Profit is directly proportional to investment.

Calculation

Mahak's capital = 18000

lalita's capital = 18000/3 × 7 = 42000

Mahak's total capital = 8 × 18000 + 4 × 14000

= 200000

lalita's total capital = 42000 × 12

Profit ratio = 200000:504000

= 25:63

The answer is 25:63

We know that Mode = 3 × Median - 2 × Mean

we have the ratio of mean and median of a distribution is 2 : 3 then

Median = 3x and Mean = 2x

Now Mode = 3 × 3x - 2 × 2x = 9x - 4x = 5x

So Mode : Mean = 5x : 2x = 5 : 2

Hence the ratio of mode and mean is 5 : 2.