SSC CGL 2019 Solved PAPER : Held on 05-March-2020 Shift-2 (Quantitative Aptitude)

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SSC CGL 2019 EXAM PAPER : Held on 05-March-2020 Shift-2 (Quantative Aptitude)

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Q.1 ABCD is a cyclic quadrilateral in which sides AD and BC are produced to meet at P, and sides DC and AB meet at Q when produced. If ∠A = 60° and ∠ABC = 72°, then ∠DPC –∠BQC = ?

1. 40°
2. 24°
3. 36°
4. 30°
Correct answer is Option 3 i.e. 36°
∠A = 60° and ∠ABC = 72°
In ΔAPB
∠PAC + ∠ABP + ∠APB = 180°
∠ABP = 180° – 60° – 72° = 48°
∠ABP = ∠DPC = 48°
As we know, In cyclic quadrilateral, sum of opposite angles are 180°.
∠ABC + ∠ADC = 180°
∠ADC = 180° – 72° = 108°
In ΔADQ
∠DAQ + ∠ADQ + ∠AQD = 180°
∠AQD = 180° – 60° – 108° = 12°
Difference between ∠DPC and ∠BQC = 48° – 12° = 36°
Q.2 The average of some numbers is 54.6. If 75% of the numbers are increased by 5.6 each, and the rest are decreased by 8.4 each, then what is the average of the numbers so obtained ?
1. 55.8
2. 55.6
3. 56.7
4. 56.3
The correct answer is option 3 i.e  56.7
Let there are 4 numbers, each value of 54.6
Average of 4 numbers = 54.6
Sum of 4 numbers = 54.6 × 4 = 218.4
According to the question,
New average of 4 numbers = 3 (54.6 + 5.6) + 1 (54.6 – 8.4) = 3 × 60.2 + 46.2 = 180.6 + 46.2 = 226.8
New average of 4 numbers = 226.8/4 = 56.7
Short Trick:
Let there are 4 numbers.
New average of 4 numbers = 54.6 + (3 × 5.6 – 1 × 8.4)/4 = 54.6 + 8.4/4 = 54.6 + 2.1 = 56.7

Q.3  To complete a certain task, X is 40% more efficient than Y, and Z is 40% less efficient than Y. Working together, they can complete the task in 21 days. Y and Z together worked for 35 days. The remaining work will be completed by X alone in:

1. 4 days
2. 5 days
3. 8 days
4. 6 days
Correct answer is option 2 i.e. 5 days
Let efficiency of y be 5 unit
Efficiency of X = 5 × (7/5) = 7
Efficiency of Z = 5 × (3/5) = 3
Total work = (5 + 7 + 3) × 21 = 315
Work done by y and Z in 35 days = (5 + 3) × 35 = 280
Remaining work = 315 – 280 = 35
∴Remaining work will be completed by X alone in = 35/7 = 5 days
Q.4 If 11 sin2 θ – cos2 θ + 4 sin θ – 4 = 0, 0° < θ < 90°, then what is the value of \frac { cos2θ+cot2θ }{ sec2θ−tan2θ }
1. \frac { 12+5\sqrt { 3 } }{ 3 }
2. \frac { 10+5\sqrt { 3 } }{ 3 }
3. \frac { 12+7\sqrt { 3 } }{ 6 }
4. \frac { 10+7\sqrt { 3 } }{ 6 }
Correct answer is Option 3 i.e. \frac { 12+7\sqrt { 3 } }{ 6 }
11 sin2 θ – cos2 θ + 4 sin θ – 4 = 0
Put θ = 30, and (satisfied)
Now,
\frac { cos2θ+cot2θ }{ sec2θ−tan2θ }
\frac { cos60+cot60 }{ sec60−tan60 }
= \frac { \frac { 1 }{ 2 } +\frac { 1 }{ \sqrt { 3 } } }{ 2-\sqrt { 3 } }
= \frac { 2+\sqrt { 3 } }{ 2\sqrt { 3 } (2-\sqrt { 3 } ) } \times \frac { 2+\sqrt { 3 } }{ 2+\sqrt { 3 } }
= \frac { 7+4\sqrt { 3 } }{ 2\sqrt { 3 } }
= \frac { 12+7\sqrt { 3 } }{ 6 }
 
Q.5 A and B together borrowed a sum of Rs. 51,750 at an interest rate of 7% p.a. compound interest in such a way that to settle the loan, A paid as much amount after three years as paid by B after 4 years from the day of borrowing. The sum (in Rs.) borrowed by A was:
1. 26,750
2. 25,000
3. 24,860
4. 25,650
Correct answer is option 1 i.e. 25,650
Short Trick:
As we know, the difference between years 4 and 3 = 1 year, then
A : B = 107 : 100
107 + 100 = 207 unit
207 unit = 51750
1 unit = 250
107 unit = 26750
Q.6 What x is added to each of 10, 16, 22 and 32, the numbers so obtained in this order are in proportion. What is the mean proportional between the numbers (x + 1) and (3x + 1)?

1. 15
2. 10
3. 12
4. 9
Correct answer is option 1 i.e. 15
If, x is added to each of 10, 16, 22 and 32, the numbers so obtained in this order are in proportion, then
According to the question
(10 + x) : (16 + x) :: (22 + x) : (32 + x)
(10 + x)/(16 + x) = (22 + x)/(32 + x)
We can solve this using cross multiplication or Using Componendo or Dividendo but hare we can solve this using equal difference between 16 – 10 = 6 and 32 – 22 = 10. For makes calculation easy and saves time.
6 × 5 = 30 and 10 × 3 = 30
Now,
5 (10 + x) = 3 (22 + x)
50 + 5x = 66 + 3x
2x = 16
x = 8
Now,
Let the mean proportional between the numbers (x + 1) and (3x + 1) be A, then
(x + 1) : A :: A : (3x + 1)
(8 + 1)/A = A/(3 × 8 + 1)
A2 = 9 × (24 + 1)
A = √(9 × 25)
A = 3 × 5 = 15
Q.7 Reema sold 48 articles for Rs. 2,160 and suffered a loss of 10%. How many articles should she sell for Rs. 2,016 to earn a profit of 12%?
1. 28
2. 36
3. 40
4. 32
Correct answer is option 2 i.e. 36
SP of 48 articles = Rs. 2160
CP of 48 articles = 2160 × (100/90) = Rs. 2400
CP of 1 article = 2400/48 = Rs. 50
New sp of 1 article = 50 × (112/100) = 56
In Rs. 56 we can sold = 1 article
In Rs. 1 we can sold = 1/56 article
In Rs. 2016 we can sold = 1/56 × 2016 = 36 articles
Q.8 If {20 x }^{ 2 } – 30x + 1 = 0,  then what is the value of { 25x }^{ 2 }+\frac { 1 }{ { 16x }^{ 2 } } 
1. 58\frac { 3 }{ 4 }
2. 53\frac { 1 }{ 2 }
3. 58\frac { 1 }{ 2 }
4. 53\frac { 3 }{ 4 }
Correct answer is option 4 i.e. 53\frac { 3 }{ 4 }
20x2 – 30x + 1 = 0
Divided by 4x
5x – 15/2 + 1/4x
5x + 1/4x = 15/2
As we know,
(5x + 1/4x)2 = 25x2 + 1/16x2 + 2 × 5x × (1/4x)
(15/2)2 = 25x2 + 1/16x2 + 10/4
25x2 + 1/16x2 = 225/4 – 10/4 = 215/4 = 53\frac { 3 }{ 4 }
Q.9 When a positive integer is divided by d, the remainder is 15. When ten times of the same number is divided by d, the remainder is 6. The least possible value of d is:

1. 9
2. 12
3. 18
4. 16
 
Correct answer is option 4 i.e  16
As given, Remainder is 15 so we can say value of d is greater than 15. Only 16 and 18 is greater than 15.
Taking, d = 16 from the options
Ten times of 15 = 150
When we divide 150 by 16 we get remainder 6 (condition satisfied). So the value of d = 16.
Q10. In ΔABC, ∠B = 90°. If points D and E are on side BC such that BD = DE = EC, then which of the following is true?
1. 5 AE2 = 2 AC2 + 3 AD2
2. 8 AE2 = 3 AC2 + 5 AD2
3. 5 AE2 = 3 AC2 + 2AD2
4. 8 AE2 = 5 AC2 + 3 AD2
Correct answer is Option 2. i.e.  8 AE2 = 3 AC2 + 5 AD2
Short trick:
In this type of question let the numerical values for savings time, calculation, and pressure.
As we know, the triplet 3, 4, and 5.
Let AB = 4 and BC = 3 and AC = 5 and BD = DE = EC = 1 cm and BE = 2 cm
Using Pythagoras theorem, AD = √17 and AE = √20
Now, check the options one by one and satisfied RHS and LHS.
From option 3.
8 AE2 = 3 AC2 + 5 AD2
8 × 20 = 3 × 25 + 5 × 17
160 = 160 (satisfied)
Detailed Method:
In ΔABC
AC2 = AB2 + BC2
AC2 = AB2 + (3BD)2
AC2 = AB2 + 9BD2   —(1)
In ΔABE
AE2 = AB2 + BE2
AE2 = AB2 + (2BD)2
AE2 = AB2 + 4BD2   —(2)
In ΔABD
AD2 = AB2 + BD2
AD2 = AB2 + BC2   —(3)
Add equation after multiplying by 3 in equation (1) and by 5 in equation (3)
3AC2 + 5AD2 = 8AB2 + 32 BD2   —(4)
Multiply by 8 in equation (2)
8AE2 = 8AB2 + 32BD2   —(5)
From equation (4) and equation (5)
8AE2 = 3AC2 + 5AD2
 
Q.11 The Value of  \frac { 3 }{ 5 } \times 1\frac { 7 }{ 8 } \div 1\frac { 1 }{ 3 } of\frac { 3 }{ 16 } -\left( 3\frac { 1 }{ 5 } \div 4\frac { 1 }{ 2 } of5\frac { 1 }{ 3 } \right) \times 2\frac { 1 }{ 2 } +\frac { 1 }{ 2 } +\frac { 1 }{ 8 } \div \frac { 1 }{ 4 } is :
1. 4\frac { 1 }{ 8 }%
2. 5\frac { 1 }{ 6 }%
3. 5\frac { 5 }{ 6 }%
4. 4\frac { 1 }{ 3 }%
Correct answer is Option 1 i.e. 16\frac { 2 }{ 3 }%
\frac { 3 }{ 5 } \times 1\frac { 7 }{ 8 } \div 1\frac { 1 }{ 3 } of\frac { 3 }{ 16 } -\left( 3\frac { 1 }{ 5 } \div 4\frac { 1 }{ 2 } of5\frac { 1 }{ 3 } \right) \times 2\frac { 1 }{ 2 } +\frac { 1 }{ 2 } +\frac { 1 }{ 8 } \div \frac { 1 }{ 4 }
= \frac { 3 }{ 5 } \times \frac { 15 }{ 8 } \div \frac { 4 }{ 3 } of\frac { 3 }{ 16 } -\left( \frac { 16 }{ 5 } \div \frac { 9 }{ 2 } of\frac { 16 }{ 3 } \right) \times \frac { 5 }{ 2 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 }
= \frac { 3 }{ 5 } \times \frac { 15 }{ 8 } \div \frac { 1 }{ 4 } -\left( \frac { 16 }{ 5 } \div 24 \right) \times \frac { 5 }{ 2 } +1
= \frac { 3 }{ 5 } \times \frac { 15 }{ 8 } \times 4-\left( \frac { 16 }{ 5 } \div \frac { 1 }{ 24 } \right) \times \frac { 5 }{ 2 } +1
= \frac { 9 }{ 2 } -\left \frac { 2 }{ 15 } \right \times \frac { 5 }{ 2 } +1
= \frac { 9 }{ 2 } -\left \frac { 1 }{ 3 } \right +1
= \frac { 27-2+6 }{ 6 }
= \frac { 31 }{ 6 }
= 5\frac { 1 }{ 6 }
Q.12 The given table represents the sale (in thousands) of cars by four companies A, B, C and D in six years. Study the table and answer the question that follows.
The total number of cars sold by company C in 2018 exceeds the average number of cars sold by company A during 2014 to 2018 by:
1. 12,000
2. 16,000
3. 15,000
4. 14,000
Correct answer is Option 4 i.e. 16000
Total number of cars sold by company A during 2014 to 2018 = 52 + 61 + 72 + 52 + 63 = 300
Average cars sold by A during 2014 to 2018 = 300/5 = 60
Cars sold by company C in 2018 = 76
The total number of cars sold by company C in 2018 exceeds the average number of cars sold by company A during 2014 to 2018 by = 76 – 60 = 16
The given table represents the sale (in thousands), so 16 = 16000
Q.13 The given table represents the sale (in thousands) of cars by four companies A, B, C and D in six years. Study the table and answer the question that follows.
The total number of cars sold by company B during 2015, 2017 and 2018 is what percentage less than the total number of cars sold by company C in 2013, 2015, 2017 and 2018?
1. 40
2. 33\frac { 1 }{ 3 }
3. 16\frac { 2 }{ 3 }
4. 50
Correct Answer is option 2 i.e. 33\frac { 1 }{ 3 }
To ease our calculation part, we can neglect 1000 part.
Total number of cars sold by company B during 2015, 2017 and 2018 = 60 + 53 + 67 = 180
Total numbers of cars sold by company C in 2013, 2015 and 2017 and 2018 = 65 + 66 + 63 + 76 = 270
Required percentage = [(270 – 180)/270] × 100 = 33\frac { 1 }{ 3 }
Q.14 The given table represents the sale (in thousands) of cars by four companies A, B, C and D in six years. Study the table and answer the question that follows.
What is the ratio of the total number of cars sold by companies A, B and D in 2017 to the total number of cars sold by all four companies in 2018?
1. 3 : 4
2. 6 : 13
3. 9 : 14
4. 18 : 23
Correct answer is Option 3 i.e.  9 : 14
Total number of cars sold by companies A, B and D in 2017 = 52 + 53 + 75 = 180
Total number of cars sold by all four companies in 2018 = 63 + 67 + 76 + 74 = 280
Required ratio = 180 : 280 = 9 : 14
Q.15 The given table represents the sale (in thousands) of cars by four companies A, B, C and D in six years. Study the table and answer the question that follows.
The total number of cars sold by companies A in 2017 and C in 2013 is what percentage of the total number of cars sold by all four companies in 2013 and 2016? (correct to one decimal place)
1. 23.3
2. 24.2
3. 25.6
4. 23.8
Correct answer is Option 1 i.e.  23.3%
Total number of cars sold by companies A in 2017 and C in 2013 = 52 + 65 = 117
Total number of cars sold by all four companies in 2013 and 2016 = 45 + 63 + 65 + 67 + 72 + 58 + 70 + 63 = 503
Required percentage = [117/503] × 100 =  23.3%
Q.16 In ΔPQR, PQ = 24 cm and ∠Q = 58°. S and T are the points on side PQ and PR, respectively, such that ∠STR = 122°.If PS = 14 cm and PT = 12 cm, then the length of RT is:
1. 15 cm
2. 16 cm
3. 16.4 cm
4. 14.8 cm
Correct answer is Option 2 i.e. 16 cm
Given, PS = 14 and PT = 12 and PQ = 24
In ΔPQR and In ΔPST
∠P = ∠P = common
∠PQR = ∠PTS = 58°
ΔPQR ∼ ΔPTS
As we know,
PR/PS = PQ/PT
PR/14 = 24/12
PR = 28
RT = PR – PT = 28 – 12 = 16 cm
Q.17 If 5 cos θ – 12 sin θ = 0, then what is the value of  \frac { 1+sinθ+cosθ }{ 1-sinθ+cosθ } is ?

1. \frac { 3 }{ 4 }

2. \frac { 5 }{ 4 }

3. \frac { 3 }{ 2 }

4. \frac { 5 }{ 2 }

Correct answer is Option 3 i.e. \frac { 3 }{ 2 } cm
5 cosθ – 12 sin θ = 0
⇒5 cos θ = 12 sinθ
⇒sinθ/cosθ = 5/12
⇒tanθ = 5/12
As we know,
Hypotenuse = H, Perpendicular = P and Base = B
tan θ = P/B = 5/12
As we know,
H2 = P2 + B2
⇒H2 = 122 + 52 = 144 + 25 = 169
⇒H = √169 = 13 cm
sinθ = P/H = 5/13 and cos θ = B/H = 12/13
Now,
(1 + sinθ + cos θ)/(1 – sinθ + cosθ)
⇒(1 + 5/13 + 12/13)/(1 – 5/13 + 12/13)
⇒(13 + 12 + 5)/(13 – 5 + 12)
⇒30/20
⇒3/2
Q18.  \frac { { x }^{ 2 }{ \left( x-4 \right) }^{ 2 } }{ { \left( x+4 \right) }^{ 2 }-4x } \div \frac { { \left( { x }^{ 2 }-4x \right) }^{ 3 } }{ { \left( x+4 \right) }^{ 2 } } \times \frac { 64-{ x }^{ 3 } }{ 16-{ x }^{ 2 } } is equal to
1. \frac { x-4 }{ x+4 }
2. \frac { x+4 }{ x(4-x) }
3. \frac { x+4 }{ x(x-4) }
4. \frac { x+4 }{ (x-4) }

Correct answer is Option 3 i.e. \frac { x+4 }{ x(x-4) }

\frac { { x }^{ 2 }{ \left( x-4 \right) }^{ 2 } }{ { \left( x+4 \right) }^{ 2 }-4x } \div \frac { { \left( { x }^{ 2 }-4x \right) }^{ 3 } }{ { \left( x+4 \right) }^{ 2 } } \times \frac { 64-{ x }^{ 3 } }{ 16-{ x }^{ 2 } }

Put x = 2

= \frac { { 2 }^{ 2 }{ \left( 2-4 \right) }^{ 2 } }{ { \left( 2+4 \right) }^{ 2 }-4\times 2 } \div \frac { { \left( 2^{ 2 }-4\times 2 \right) }^{ 3 } }{ { \left( 2+4 \right) }^{ 2 } } \times \frac { 64-{ 2 }^{ 3 } }{ 16-{ 2 }^{ 2 } }

= \frac { 4×4 }{ 36−8 } \div \frac { −64 }{ { 36 } } \times \frac { 64-8 }{ 16-4 }

= \frac { 16 }{ 28 } \times \frac { 36 }{ { -64 } } \times \frac { 56 }{ 12 }

= \frac { -3 }{ 2 }

From option 3

\frac { x+4 }{ x(x-4) }

\frac { 2+4 }{ 2(2-4) }

= \frac { -6 }{ 4 }

= \frac { -3 }{ 2 }

Q.19 Diameter AB of a circle with centre O is produced to a point P such that PO = 16.8 cm. PQR is a secant that intersects the circle at Q and R such that PQ = 12 cm and PR = 19.2 cm. The length of AB (in cm) is:
1. 15.8
2. 14.4
3. 15.2
4. 14.2
Correct answer is Option 2 i.e. 31°
Given, PO = 16.8, PQ = 12 and PR = 19.2
From the following figure
Let the radius of the circle AO = OB = x, then
PA = (16.8 + x) and PB = (16.8 – x)
As we know,
PB × PA = PQ × PR
(16.8 – x) (16.8 + x) = 12 × 19.2
16.82 – x2 = 12 × 19.2
x2 = 16.82 – 12 × 19.2 = 122 (1.42 – 1.6) = 144 × (1.96 – 1.60)
x = √(144 × 0.36)
x = 7.2
Hence, 2x = 14.4
∴AB = 14.4
For easy calculation
Given,
PO = 16.8, PQ = 12 and PR = 19.2
Divided by 12, then
PO = 1.4, PQ = 1 and PR = 1.6
Let the radius of the circle AO = OB = x, then
PA = (1.4 + x) and PB = (1.4 – x)
As we know,
PB × PA = PQ × PR
(1.4 – x) (1.4 + x) = 1 × 1.6
1.96 – x2 = 1.6
x2 = 0.36
x = √0.36
x = 0.6
2x = 1.2
AB = 1.2
Multiply by 12 in 1.2 and get correct answer = 1.2 × 12 = 14.4
Q.20 What is the value of \frac { cosec(78°−θ)−sec(12°+θ)−tan(67°+θ)+cot(23°−θ) }{ tan13°tan37°tan45°tan53°tan77° } ?

1. 2
2. 1
3. 0
4. -1
Correct answer is Option 3 i.e. 0
[cosec (78 – θ) – sec (12 + θ) – tan (67 + θ) + cot (23 – θ)]/[tan13 cot 37 tan 45 tan53 tan77]
As we know,
cosec (90 – θ) = sec θ and tanθ cotθ = 1 and tan 45 = 1 and tan θ tan (90 – θ) = 1
⇒ [cosec (78 – θ) – cosec {90 – (12 + θ)} – tan (67 + θ) + tan {90 – (23 – θ)}]/1
⇒ [cosec (78 – θ) – cosec (78 – θ) – tan (67 + θ) + tan (67 + θ)]
⇒ 0
Q.21 A and B spend 60% and 75% of their incomes, respectively. If the savings of A are 20% more than that of B, then by what percentage is the income of A less than the income of B?
1. 25
2. 15
3. 20
4. 10
Correct answer is Option 1  i.e. 25
Using option 1
let income of B be 40
Income of A = 30
Savings of A = 30 × (40/100) = 12
Savings of B = 40 × (25/100) = 10
Savings of A is more than B by = (12 – 10)/10 × 100 = 20% (satisfied)
Detailed Method:
Let the income of A and B be Rs. x and y respectively.
Savings of A = 40% of x
Savings of B = 25% of y
According to the question
(40% of x) = (25% of y) × (120/100)
x × 40% = y × 25% × 6/5
x : y = 3 : 4
Let income of A = 3 and income of B = 4
Income of A is less than that of B by = [(4 – 3)/4] × 100 = 25%
Q.22 A boat can go 3.6 km upstream and 5.4 km downstream in 54 minutes, while it can go 5.4 km upstream and 3.6 km downstream in 58.5 minutes. The time (in minutes) taken by the boat in going 10 km downstream is:
1. 50
2. 48
3. 45
4. 54
 
Correct answer is Option 1 i.e. 50
Short Trick:
U = upstream and D = downstream
3.6(U) + 5.4(D) = 54 and 5.4(U) + 3.6(D) = 58.5
Divide by 18
2U + 3D = 30 and 3U + 2D = 32.5
Now, Multiply by 3 in (1) and by 2 in (2), and then subtract (2) from (1), then
5D = 25 min
∴10D = 50 min
Detailed Method:
Let the speed of boat be x km/hr and speed of current be y km/hr
Upstream speed = (x – y) km/hr
Downstream speed = (x + y) km/hr
According to the question
3.6/(x – y) + 5.4/(x + y) = 54/60
5.4/(x – y) + 3.6/(x + y) = 58.5/60
Let 1/(x – y) = a and 1/(x + y) = b, then
3.6a + 5.4b = 54/60   —(1)
5.4a + 3.6b = 58.5/60   —(2)
Add equation (1) and equation (2)
9 (a + b) = 112.5/60
⇒a + b = 12.5/60
⇒a + b = 5/24   —(3)
Subtract equation (2) from equation (1)
1.8 (b – a) = -4.5/60
⇒b – a = -2.5/60
⇒b – a = -1/24   —(4)
Add equation (3) and equation (4)
2b = 4/24
⇒b = 1/12
⇒(x + y) = 12 km/hr
Downstream speed = 12 km/hr
∴Time taken to cover 10 km distance with downstream = 10/12 hr or (10/12) × 60 = 50 min
Q.23 If x4 + x2y2 + y4 = 273 and x2 – xy + y2 = 13, then the value of xy is:
1. 4
2. 10
3. 6
4. 8
Corrrect answer is Option 1 i.e. 4
x2 – xy + y2 = 13   —(1)
x4 + x2y2 + y2 = 273 
As we know,
(x2 – xy + y2)(x2 + xy + y2) = x4 + x2y2 + y2
13 × (x2 + xy + y2) = 273
(x2 + xy + y2) = 273/13 = 21   —-(2)
From equation (1) and equation (2)
2xy = 8
xy = 4
Q.24 The marked price of an item is 25% above its cost price. A shopkeeper sells it, allowing a discount of x% on the marked price. If he incurs a loss of 8%, then the value of x is:
1. 25.2%
2. 26.8%
3. 25.6%
4. 26.4%
Correct answer is Option 4 i.e. 26.4%
Let CP of the article be Rs. 100
MP of the article = 100 × (125/100) = Rs. 125
SP of the article = 100 × (92/100) = Rs. 92
Discount = 125 – 92 = 33
Discount percentage = 33/125 × 100 = 26.4%
Short Trick:
-8 = 25 – x – (25x/100)
-8 = 25 – x – x/5
5x/4 = 33
x = 132/5 = 26.4%
Q.25 If the radius of a right circular cylinder is decreased by 10%, and the height is increased by 20%, then the percentage increase/decrease in its volume is:

1. increase by 1.8%
2. decrease by 2.8%
3. decrease by 1.8%
4. increase by 2.8%
Correct answer is Option 2  i.e. decrease by 2.8%

Ratio of volume before and after = 500 : 486
Again multiply by 2 in = 1000 : 972
Volume decreased by = 1000 – 972 = 28
As we know, 28 = 2.8%
Detailed Solution:
Let the radius of the cylinder be r cm and height of the cylinder be h cm,
As we know,
Volume of the cylinder = πr2h
New radius of the cylinder = r × (90/100) = 0.9r cm
New height of the cylinder = h × (120/100) = 1.2h cm
New volume of the cylinder = π × (0.9r)2 × 1.2h = 0.81r2 × 1.2h × π = 0.972rhπ
Required percentage = [(πr2h – 0.972πrh)/πrh] × 100 = 2.8%

Download Complete SSC CGL 2019 Paper Held on 03 March 2020 Shift 1 paper here Click Download below.

manish aggarwal
Founder of edumo, SSC CGL Aspirant, Educator, BCA Graduate
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