[SSC CGL 2020]*SSC CGL Quantitative Aptitude 24 August 2021 : Shift 1 Question and Solutions in English

The vertices of a Δ ABC lie on a circle with centre O. AO is produced to meet the circle at the point P. D is a point on BC such that AD ⊥ BC. If ∠B = 68° and ∠C = 52°, then the measure of ∠DAP is:

12°
18°
16°
28°

Answer (Detailed Solution & Explanation Below)

Option 3 : 16°

Given:

AD ⊥ BC

∠B = 68°

∠C = 52°

Calculation:

Join O to B where point O is the circum centre of the triangle.

⇒ ∠BOA = 2∠BCA

⇒ 2 × 52°

⇒ 104°

In ΔBOA

⇒ OB = OA [radius of the circle]

⇒ ∠OBA = ∠OAB

⇒ ∠AOB + ∠OAB + ∠OBA = 180°

⇒ 2∠OAB = (180° – 104°)

⇒ 2∠OAB = 76°

⇒ ∠OAB = (76/2)

⇒ ∠OAB = 38°

In ΔABD

⇒ ∠ABD + ∠BDA + ∠DAB = 180°

⇒ ∠DAB = (180° – 90° – 68°)

⇒ ∠DAB = 22°

Now,

∠DAP = ∠BAO – ∠BAD

⇒ ∠DAP = (38° – 22°)

⇒ ∠DAP = 16°

So, the required value is 16°

Question. 16.

A can complete a work in 11(1)/(2) days. B is 25% more efficient than A and C is 50% efficient than B.Working together A. B and C will complete the same work.

5 days
4 days
3 days
8 days

Answer (Detailed Solution & Explanation Below)

Option 2 : 4 days

Given:

Time taken to complete a work by A = days = 23/2 days

B is 25% more efficient than A = (25/100) = 1/4

Efficiency of A = 4 days

Efficiency of B = (4 + 1) = 5 days

C is 50% efficient than B = (50/100) = 1/2

Efficiency of C = (5/2) = 2.5 days

Formula used:

Work = Efficiency × Time

Calculation:

Total work = (23/2 × 4) = 46 days

Total time taken by A, B and C together = [46/(4 + 5 + 2.5)] days

⇒ (46/11.5) days

⇒ 4 days

∴ The required time is 4 days

Shortcut Trick:-

Efficiency of them

So, total work = 8 × 23/2 = 92 units

Total time taken by A, B and C together = [92/(8 + 10 + 5)] = 4 days

Question. 17.

The average of 23 numbers is 51.The average of first 12 numbers is 49 and the average of last 12 numbers is 54.If the twelfth number is removed then the average of the remaining numbers (correct to two decimal places) is:

53.25
50.45
51.75
52.65

Answer (Detailed Solution & Explanation Below)

Option 2 : 50.45

Given:

The average of 23 numbers = 51

The average of first 12 numbers = 49

The average of last 12 numbers = 54

Formula used:

Average = sum of observations/Number of observations

Calculation:

The sum of 23 numbers = (23 × 51) = 1173

The sum of first 12 numbers = (12 × 49) = 588

The sum of last 12 numbers = (12 × 54) = 648

Now,

The twelfth number = [(588 + 648) – 1173]

⇒ (1236 – 1173)

⇒ 63

The 12th number = 63

Now, If the twelfth number is removed,

The sum of remaining numbers = (1173 – 63) = 1110

The average of remaining number = (1110/22)

⇒ 50.45

∴ The required average number is 50.45

Question. 18.

The total mumber of students in a school is 1400 out of which 35% of the students are girls and the rest are boys.If 50% of the boys and 90% of the girls passed in an ammal examination then the percentage of the students who failed is:

16.5
21.5
17.4
15.8

Answer (Detailed Solution & Explanation Below)

Option 1 : 16.5

Given:

The total number of students in a school = 1400

Number of girls = 35%

Number of boys who passed an annual examination = 80%

Number of girls who passed an annual examination = 90%

Calculation:

The total number of students in a school = 1400

Number of girls = (1400 × 35/100)

⇒ 490

Number of boys = (1400 – 490) = 910

Number of boys who passed an annual examination = (80% of 910)

⇒ (80/100 × 910)

⇒ 728

Number of boys who failed the annual examination = (910 – 728) = 182

Number of girls who passed an annual examination = (90% of 490)

⇒ (90/100 × 490)

⇒ 441

Number of girls who failed the annual examination = (490 – 441) = 49

Now,

Total number of failed students = (182 + 49) = 231

Required percentage of failed students = (231/1400 × 100)

⇒ (231/14)

⇒ 16.5%

∴ The required percentage of failed students is 16.5

Question. 19.

If x + 1/x = 7, then find the value of x2 + 1/x2

Answer (Detailed Solution & Explanation Below)

Option 1 : 47

Given:

x + 1/x = 7

Formula used:

x + 1/x = k

⇒ x2 + 1/x2 = (k)2 – 2

Calculation:

x + 1/x = 7

⇒ x2 + 1/x2 = (7)2 – 2

= 49 – 2 = 47

Question. 20.

In a triangle ABC, length of the side AC is 4 cm more than 2 times the length of the side AB. Length of the side BC is 4 cm less than the three times the length of the side AB. If the perimeter of ΔABC is 60 cm, then its area (in cm2) is:

Answer (Detailed Solution & Explanation Below)

Option 4 : 120

Given:

Length of the side AC is 4 cm more than 2 times the length of the side AB

Length of the side BC is 4 cm less than the three times the length of the side AB.

The perimeter of ΔABC = 60 cm

Formula used:

Area of triangle = 1/2 × base × height

Calculation:

Let the side of AB be x cm

Then,

Length of the side AC is 4 cm more than 2 times the length of the side AB = (2x + 4) cm

Length of the side BC is 4 cm less than the three times the length of the side AB. = (3x – 4) cm

Perimeter of ΔABC = 60 cm

⇒ (x + 2x + 4 + 3x – 4) = 60

⇒ 6x = 60

⇒ x = 10 cm

Now,

Side of AB = x = 10 cm

Side of AC = (2x + 4) = (2 × 10 + 4) = 24 cm

Side of BC = (3x – 4) = (3 × 10 – 4) = 26 cm

We know that 10, 24, and 26 is a triplet,

⇒ We can say that the given triangle is a Right angle triangle.

Now,

⇒ Area of triangle = 1/2 × base × height

Area of triangle = (1/2 × 10 × 24)

⇒ (5 × 24)

⇒ 120 cm2

∴ The required area is 120 cm2

Question. 21.

If 2x + 3y + 1 = 0, then what is the value of (8x3 + 8 + 27y3 - 18xy)?

-7
7
-9
9

Answer (Detailed Solution & Explanation Below)

Option 2 : 7

Given:

2x + 3y = 1

Formula used:

(a + b)3 = a3 + b3 + 3ab(a + b)

Calculation:

According to the question

2x + 3y + 1 = 0

⇒ 2x + 3y = -1

Now,

(2x + 3y)3 = (2x)3 + (3y)3 + 3 × 2x × 3y (-1) = -1

⇒ 8x3 + 27y3 – 18xy = -1

Now,

The value of (8x3 + 8 + 27y3 – 18xy) = (8 – 1)

⇒ 7

∴ The required value is 7

Question. 22.

A sum of Rs. 9500 amounts to Rs. 11495 in 2 years at a certain rate percent per annum, interest compounded yearly. What is the simple interest (in Rs. ) on the same sum for the same time and double the rate?

3990
3420
4560
3800

Answer (Detailed Solution & Explanation Below)

Option 4 : 3800

Given:

Amount = Rs. 11495

Principal = Rs. 9500

Time = 2 years

Formula used:

A = P(1 + r/100)t

SI = (P × R × T)/100

Calculation:

A = P(1 + r/100)t

⇒ 11495 = 9500(1 + r/100)2

⇒ (1 + r/100)2 = (11495/9500)

⇒ (1 + r/100)2 = (121/100)

⇒ (1 + r/100)2 = (11/10)2

⇒ r = 10%

Now,

According to the question

Principal = Rs. 9500

Rate = double the rate = (2 × 10) = 20%

Time = 2 years

SI = (P × R × T)/100

⇒ Rs. (9500 × 20 × 2)/100

⇒ Rs. (95 × 20 × 2)

⇒ Rs. 3800

∴ The required SI is Rs. 3800

Question. 23.

Table shows income (in Rs.) recieved by 4 employee of a company during the month of December 2020 and all their income sources.
Source Amit Suresh Nitin Varun
Salary 35000 38500 29000 42000
Arrears 6000 6300 5000 7500
Bonus 1000 1100 1000 1240
Overtime 1800 1950 1400 1500

By what percent is the bonus of Varun less than the bonus of Amit and Nitin taken together?

45
40.9
48
38

Answer (Detailed Solution & Explanation Below)

Option 4 : 38

Calculation:

Bonus amount for Varun = Rs.1240

Bouns amount for Amit and Nitin = 1000 + 1000 = Rs.2000

Required % = (2000 - 1240)/2000 × 100 = 38%

∴ Bonus of Varun is 38% less than the bonus of Amit and Nitin taken together.

Question. 24.

Simplify: $\rm \sec^2 \alpha\left(1+\frac{1}{cosec\ \alpha}\right)\left(1-\frac{1}{cosec\ \alpha}\right)$

sin2 α
1
-1
tan4 α

Answer (Detailed Solution & Explanation Below)

Option 2 : 1

Given:

$\rm \sec^2 α\left(1+\frac{1}{cosec\ α}\right)\left(1-\frac{1}{cosec\ α}\right)$

Formula used:

(a2 – b2) = (a + b)(a – b)

sec2α = 1/cos2α

1/cosecα = sinα

1 – sin2α = cos2α

Calculation:

$\rm \sec^2 α\left(1+\frac{1}{cosec\ α}\right)\left(1-\frac{1}{cosec\ α}\right)$

⇒ 1/cos2α (1 + sinα)(1 – sinα)

⇒ 1/cos2α (12 – sin2α)

⇒ 1/cos2α (1 – sin2α)

⇒ 1/cos2α × (cos2α)

⇒ 1

∴ The required value is 1

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Question. 25.

The value of 3(1)/(5)-:4(1)/(2) of 5(1)/(3)-(1)/(8)-:(1)/(2) of (1)/(4)+(1)/(4)((1)/(2)-:(1)/(8)times(1)/(4)) is :