SSC CGL 2019 EXAM PAPER : Held on 05-March-2020 Shift-1 (Quantitative Aptitude)

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SSC CGL 2019 EXAM PAPER : Held on 05-March-2020 Shift-1 (Quantative Aptitude)

Table of Contents

Q.1 In an examination in which the full marks were 500, A scored 25% more marks than B, B scored 60% more marks than C and C scored 20% less marks than D. If A scored 80% marks, then the percentage of marks obtained by D is:

1. 65%
2. 60%
3. 50%
4. 54%
Correct answer is Option 3 i.e.50%
Maximum marks of the examination = 500
A scored = 500 × (80/100) = 400
B scored = 400 × (100/125) = 320
C scored = 320 × (100/160) = 200
D scored = 200 × (100/80) = 250
Percentage of D = 250/500 × 100 = 50%
Other method:
Ratio of A to B = 5 : 4
Ratio of B to C = 8 : 5
Ratio of C to D = 4 : 5
Taking only ratio A to D = 160 : 100
Divide by 2 only, then A : D = 80 : 50
If A got 80%, then D got 50% 
Q.2 The given table represents the exports (in Rs. Crores) of four items A, B, C and D over a period of six years. Study the table and answer the question that follows.
In which year, the exports of item D were 1.4 times the average exports of item B during six years?
1. 2013
2. 2012
3. 2011
4. 2014
The correct answer is option 1 i.e 2013
Average export of items B during six years = (128 + 134 + 138 + 169 + 182 + 209)/6 = 960/6 = 160
⇒ 1.4 of 160 = 224
In 2013, the exports of item D were 1.4 times the average exports of item B during six years.

Q.3 If { x }^{ 2 }-2\sqrt { 5 } x+1=0 

then what is the value of  { x }^{ 5 }+\frac { 1 }{ { x }^{ 5 } } 

1. 610\sqrt { 5 }
2. 406\sqrt { 5 }
3. 408\sqrt { 5 }
4. 612\sqrt { 5 }
Correct answer is option 1 i.e. 610\sqrt { 5 }
x2 – 2√5 x + 1 = 0
Divide by x
x – 2√5 + 1/x = 0
x + 1/x = 2 √5      —— (1)
Squaring on both sides equation (1)
(x + 1/x)2 = (2 √5)2
⇒ x2 + 1/x2 + 2 = 20
⇒ x2 + 1/x2 = 20 – 2
(x2 + 1/x2) = 18      —— (2)
Cube of equation (1)
(x + 1/x)3 = (2 √5)3
⇒ x3 + 1/x3 + 3 (x + 1/x) = 40 √5
⇒ x3 + 1/x3 + 3 (2√5) = 40√5
⇒ x3 + 1/x3 = 40√5 – 6√5 = 34√5
⇒ x3 + 1/x3 = 34√5      —— (3)
Multiply by equation (2) and equation (3)
(x2 + 1/x2) (x3 + 1/x3) = 18 × 34√5
⇒ x5 + 1/x + x + 1/x5 = 612√5
⇒ x5 + 1/x5 = 612√5 – (x + 1/x)
⇒ x5 + 1/x5 = 612 √5 – 2 √5
⇒ x5 + 1/x5 = 610 √5
Short trick:
As we know,
If x + 1/x = a, then
⇒ x5 + 1/x5 = (a2 – 2)(a3 – 3a) – a
If x + 1/x = 2√5, then
⇒ x5 + 1/x5 = [(2√5)2 – 2)][(2√5)3 – 3 × 2√5)] – 2√5
⇒ x5 + 1/x5 = (20 – 2) (40√5 – 6√5) – 2√5 = 18 × 34√5 – 2√5 = 610√5
Q.4 Sudha sold an article to Renu for Rs. 576 at a loss of 20% Renu spent a sum of Rs. 224 on its transportation and sold it to Raghu at a price which would have given Sudha a profit of 24%. The percentage of gain for Renu is:
1. 13.2%
2. 10.5%
3. 12.9%
4. 11.6%
Correct answer is Option 4 i.e. 11.6%
Sudha sold an article to Renu for = Rs. 576
Cost price of the article for Sudha = Rs. 576 × (100/80) = Rs. 720
Final cost price of the article for Renu = 576 + 224 = Rs. 800
Profit gain on the article for Renu = Rs. 720 × (124/100) = Rs. 892.8
Profit percentage on the article for Renu = (92.8/800) × 100 = 11.6%
Q.5 If the nine-digit number 708x6y8z9 is divisible by 99 then what is the value of x + y + z?
1. 5
2. 27
3. 16
4. 9
Correct answer is option 3 i.e. 16
Divisibility law of 11: If the difference of the alternating sum of digits of the number is a multiple of 11 (e.g. 2343 is divisible by 11 because 2 – 3 + 4 – 3 = 0, which is a multiple of 11).
Divisibility  law of 9 ⇒ A number is divisible by 9 if the sum of its digit is divisible by 9.
The nine–digit number 708x6y8z9 is divisible by 99. So we can say number also divisible by 9 and 11.
7 + 0 + 8 + x + 6 + y + 8 + z + 9
⇒ 38 + x + y + z
Put the value of (x + y + z) from the option one by one.
If we put (x + y + z) = 16, then
⇒ 38 + 16
⇒ 54
As we know, 54 divisible by 9.
Q.6 The ratio of the ages of A and B 8 years ago was 2 : 3. Four years ago, the ratio of their ages was 5 : 7. What will be the ratio of their ages 8 years from now?

1. 4 : 5
2. 5 : 6
3. 7 : 8
4. 3 : 4
Correct answer is option 1 i.e. 4 : 5
The ratio of the ages of A and B, 8 years ago = 2x : 3x
According to the question
(2x + 4)/(3x + 4) = 5/7
⇒ 7 (2x + 4) = 5 (3x + 4)
⇒ 14x + 28 = 15x + 20
⇒ 15x – 14x = 28 – 20
⇒ x = 8
8 years ago-age of A = 2 × 8 = 16
8 years after-age of A = 16 + 8 + 8 = 32
8 years ago-age of B = 3 × 8 = 24
8 years after-age of B = 24 + 8 + 8 = 40
Ratio of ages of A to B after 8 years = 32 : 40 = 4 : 5
Short Trick:
Ratio of ages of A to B, 8 years ago = 2 : 3 or 4 : 6.
1 unit = 4 years
3 unit = 12 years
∴ Ratio of ages A to B, 8 years after = 4 : 5
Q.7 The given table represents the exports (in Rs. Crores) of four items A, B, C and D over a period of six years. Study the table and answer the question that follows.
The total exports of item A from 2012 to 2014 is what percent less than the total exports of all the four items in 2015? (Correct to one decimal place)
1. 16.7%
2. 15.2%
3. 14.3%
4. 13.8%
Correct answer is option 3 i.e. 14.3%
Total exports of item A from 2012 to 2014 = 225 + 370 + 425 =  1020
Total exports of all the four items in 2015 = 400 + 209 + 306 + 275 = 1190
∴ Required percentage = (1190 – 1020)/1190 × 100 = 14.3%
Q.8 The given table represents the exports (in Rs. Crores) of four items A, B, C and D over a period of six years. Study the table and answer the question that follows.
What is the ratio of the total exports of item A in 2014 and 2015 to the total export of item C in 2011 and 2015?
1. 7 : 5
2. 3 : 2
3. 4 : 3
4. 5 : 4
Correct answer is option 2 i.e. 3:2
Total exports of items A in 2014 and 2015 = 425 + 400 = 825
Total export of item C in 2011 and 2015 = 244 + 306 = 550
∴ Required ratio = 825 : 550 = 3 : 2
Q.9 The average of 24 numbers is 56. The average of the first 10 numbers is 71.7 and that of the next 11 number is 42. Then next three numbers (i.e 22nd, 23rd, and 24th) are in the ratio 1/2 : 1/3 : 5/12 What is the average of the 22nd and 24th numbers?

1. 58
2. 49.5
3. 55
4. 60.5
 
Correct answer is option 4 i.e  60.5
Average of 24 numbers = 56
Sum of 24 numbers = 56 × 24 = 1344
Average of first 10 numbers = 71.7
Sum of first 10 numbers = 71.7 × 10 = 717
Average of next 11 numbers = 42
Sum of next 11 numbers = 42 × 11 = 462
Sum of 22nd, 23rd and 24th = 1344 – 717 – 462 = 165
Ratio of 22nd, 23rd and 24th = 1/2 : 1/3 : 5/12 = 12/2 : 12/3 : (12 × 5)/12 = 6 : 4 : 5
Ratio of 22nd, 23rd and 24th = 6x : 4x : 5x
6x + 4x + 5x = 165
⇒ 15x = 165
⇒ x = 11
22nd number = 6x = 6 × 11 = 66
24th number = 5x = 5 × 11 = 55
∴ Average of 22nd and 24th number = (66 + 55)/2 = 121/2 = 60.5
Short Trick:
Average of 24 numbers = 56
Deviation on first 10 numbers = (56 – 71.7) × 10 = –157
Deviation on next 11 numbers = (56 – 42) 11 = 154
Sum of 22nd, 23rd and 24th = 56 × 3 – 157 + 154 = 168 – 3 = 165
Ratio of 22nd, 23rd and 24th = 1/2 : 1/3 : 5/12 = 12/2 : 12/3 : (12 × 5)/12 = 6 : 4 : 5
Average of 22nd and 24th number = (6 + 5)/2 = 5.5
6 + 4 + 5 = 15 unit
⇒ 15 unit = 165
⇒ 1 unit = 11
⇒ 5.5 unit = 60.5
Q10. If P =\frac { { x }^{ 3 }+{ y }^{ 3 } }{ { (x-y) }^{ 2 }+3xy } , Q= \frac { { (x+y) }^{ 2 }-3xy }{ { x }^{ 3 }-{ y }^{ 3 } } and R =\frac { { (x+y) }^{ 2 }+{ (x-y) }^{ 2 }}{ { x }^{ 2 }-{ y }^{ 2 } } , then what is the value of(P ÷ Q) × R?
1. 2({ x }^{ 2 }+{ y }^{ 2 })
2. { x }^{ 2 }+{ y }^{ 2 }
3. 4xy
4. 2xy
Correct answer is Option 1. i.e. 2({ x }^{ 2 }+{ y }^{ 2 })
\frac { { x }^{ 3 }+{ y }^{ 3 } }{ { (x-y) }^{ 2 }+3xy } \div \frac { { (x+y) }^{ 2 }-3xy }{ { x }^{ 3 }-{ y }^{ 3 } }\times \frac { { (x+y) }^{ 2 }+{ (x-y) }^{ 2 }}{ { x }^{ 2 }-{ y }^{ 2 } }
Put x = 1 and y = 0, then
= \frac { 1+0 }{ 1+0 } \div \frac { 1 }{ 1 } \times \frac { 1+1 }{ 1 }
1 ÷ 1×2
⇒ 1 × 2
⇒ 2
From option 1.
⇒ 2 (x2 + y2)
⇒ 2 × (1 + 0)
⇒ 2 × 1
2 (satisfied)
 
Q.11 A shopkeeper bought 80 kg of rice at a discount of 10% Besides 1kg rice was offered free to him on the purchase of every 20 kg rice. If he sells the rice at the marked price, his profit percentage will be:
1. 16\frac { 2 }{ 3 }%
2. 15\frac { 1 }{ 3 }%
3. 15\frac { 3 }{ 7 }%
4. 14\frac { 2 }{ 7 }%
Correct answer is Option 1 i.e. 16\frac { 2 }{ 3 }%
Let SP of 1 kg rice = Rs. 1
SP of 80 kg rice = Rs. 80
CP of 80 kg rice = 80 × (90/100) = Rs. 72
Rice got free on 80 kg = 80/20 = 4 kg
Total rice = 80 + 4 = 84 kg
Total SP of total rice = Rs. 84 kg.
Profit = 84 – 72 = Rs. 12
∴ Profit percentage = [12/72] × 100 =16\frac { 2 }{ 3 }%
Q.12 D is the midpoint of side BC of ΔABC. Point E lies on AC such that CE = AC/3. BE and AD intersect at G. What is AG/GD ?
1. 5 : 2
2. 8 : 3
3. 3 : 1
4. 4 ∶ 1
Correct answer is Option 4 i.e. 4:1
Given:
D is the midpoint of side BC of ΔABC 
CE = AC/3 
BE and AD intersect at G 
Calculations:
Let DP is parallel to BE 
In ΔBCE and ΔDCP 
∠CEB = ∠CPD (The corresponding angles between two parallel lines are equal)
∠CBE = ∠CDP ( The corresponding angles between two parallel lines are equal)
By AA (Rule of similarity), ΔBCE ∼ ΔDCP 
∴ (BC/DC) = (CE/CP) 
⇒ (BD + DC)/DC = (CP + PE)/CP
⇒ BD/DC + 1 = 1 + PE/CP
⇒ 1/1 = PE/CP
⇒ PE = CP = CE/2 
⇒ PE/CE = 1/2
In ΔADP and ΔAGE 
∠ADP = ∠AGE 
∠APD = ∠AEG  
By AA (Rule of similarity),ΔADP ∼ ΔAGE  
∴ (AD/AG) = (AP/AE)
⇒ (AG + GD)/AG = (AE + PE)/AE 
⇒ 1 + GD/AG = 1 + PE/AE 
⇒ GD/AG = PE/AE 
⇒ AG/GD = AE/PE
⇒ AG/GD = AE/(CE/2)
⇒ AG/GD = 2 × AE/CE 
⇒ AG/GD = 2 × 2/1 
⇒ AG/GD = 4 ∶ 1 
∴ AG/GD = 4 ∶ 1

            Alternate Method:

Given,
BD : DC = 1 : 1
AC : CE = 3 : 1
AE : CE = 2 : 1
From the following figure balancing sides of the triangle, then
LCM of 2, 1, 1 and 1 is 2
∴ AG : GD = 4 : 1
 
Q.13 Two chords AB and CD of a circle with centre O intersect each other at P. fi ∠ APC = 95° and ∠ AOD = 110°.∠BOC is:
1. 60°
2. 70°
3. 65°
4. 55°
Correct Answer is option 1 i.e.60°
∠AOD = 110 and ∠APC = 95
⇒ ∠AOD = 2∠ACD
As we know,
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∠ACD = 110/2 = 55
In ΔACP
∠CAP + ∠APC + ∠PCA = 180
⇒ ∠CAP + 95 + 55 = 180
⇒ ∠CAP = 180 – 95 – 55 = 30
⇒ ∠CAB = ∠BOC/2
⇒ ∠BOC = 30 × 2 = 60
Q.14 A train takes 2\frac { 1 }{ 2 }  hours less for a journey of 300 km, if its speed is increased by 20 km/h from its usual speed. How much time will it take to cover a distance of 192 km at its usual speed?
1. 3 hours
2. 2.4 hours
3. 4.8 hours
4. 6 hours
Correct answer is Option 3 i.e. 4.8 hours
Short trick:
Let the speed of train be x km/hr, then
⇒ [x (x + 20)/20] × 5/2 = 300
⇒ x (x + 20) = 300  × 2/5  × 20
⇒ x (x + 20) = 40 × 60
Put x = 40, then equation satisfied.
Time taken to cover 192 km = 192/40 = 4.8 hrs
Detailed Method:
Let speed of train be x km/hr
According to the question
[300/x] – [300/(x + 20)] = 5/2
⇒ 300 [x + 20 – x]/[x (x + 20) = 5/2
⇒ 6000/(x2 + 20x) = 5/2
⇒ 12000 = 5x2 + 100x
⇒ x2 + 20x – 2400 = 0
⇒ x2 + 60x – 40x – 2400 = 0
⇒ x (x + 60) – 40 (x + 60) = 0
⇒ (x + 60) (x – 40) = 0
Taking,
⇒ (x – 40) = 0
⇒ x = 40 km/hr
Speed of the train = 40 km/hr
∴ Time taken to cover 192 km by train = 192/40 = 4.8 hrs
Q.15 f 12cos2 θ – 2sin2θ + 3cosθ = 3, 0° < θ < 90°, then what is the value of \frac { cosesθ+secθ }{ tanθ+cotθ } ?
1. \frac { 4+\sqrt { 3 } }{ 4 }
2. \frac { 1+2\sqrt { 2 } }{ 2 }
3. \frac { 1+\sqrt { 3 } }{ 2 }
4. \frac { 2+\sqrt { 3 } }{ 4 }
Correct answer is Option 3 i.e. \frac { 1+\sqrt { 3 } }{ 2 }
12cos2 θ – 2sin2θ + 3cosθ = 3
⇒ 12 cos2 θ – 2 (1 – cos2 θ) + 3 cos θ = 3
⇒ 12 cos2 θ – 2 + 2 cos2 θ + 3 cos θ = 3
⇒ 14 cos2 θ + 3 cos θ – 5 = 0
⇒ 14 cos2 θ + 10 cos θ – 7 cos θ – 5 = 0
⇒ 2 cos θ (7 cos θ + 5) – 1 (7cos θ + 5) = 0
⇒ (2 cos θ – 1) (7 cos θ + 5) = 0
Taking,
(2 cos θ – 1) = 0
⇒ cos θ = 1/2
⇒ cos θ = cos 60
⇒ θ = 60
\frac { cosesθ+secθ }{ tanθ+cotθ }
= \frac { coses60+sec60 }{ tan60+cot60 }
= \frac { 2+\frac { 2 }{ \sqrt { 3 } } }{ \sqrt { 3 } +\frac { 1 }{ \sqrt { 3 } } }
= \frac { \frac { 2\sqrt { 3 } +2 }{ \sqrt { 3 } } }{ \frac { 4 }{ \sqrt { 3 } } }
= \frac { \sqrt { 3 } +1 }{ 2 }
 
Q.16 If 16a4 + 36a2b2 + 81b4 = 91 and 4a2 + 9b2 – 6ab = 13, then what is the value of 3ab?
1. \frac { 3 }{2 }
2. -3
3. -\frac { 3 }{2}
4. 5
Correct answer is Option 3 i.e. -\frac { 3 }{2}
(x4 + x2y2 + y4) = (x2 + xy + y2) (x2 – xy + y2)
⇒ 4a2 + 9b2 – 6ab = 13 
⇒ [4a2 – 6ab + 9b2] = 13     —— (1)
⇒ 16a4 + 36a2b2 + 81b4 = 91
⇒ [4a2 + 6ab + 9b2] [4a2 – 6ab + 9b2] = 91
⇒ [4a2 + 6ab + 9b2] = 91/13
⇒ [4a2 + 6ab + 9b2] = 7     —— (2)
Subtract equation (2) from equation (1)
12ab = 7 – 13
⇒ ab = –6/12
⇒ 3ab = –3/2
Q.17 In ΔABC, ∠B = 90°, AB = 5 cm and BC = 12 cm the bisector of ∠A meets BC at D. What is the length of AD?

1. \frac { 2 }{ 3 } \sqrt { 13 }cm

2. \frac { 4 }{ 3 } \sqrt { 13 }cm

3. 2\sqrt { 13 } cm

4. \frac { 5\sqrt { 13 } }{ 3 } cm

Correct answer is Option 4 i.e.  \frac { 5\sqrt { 13 } }{ 3 } cm
In right triangle ABC
AC2 = AB2 + BC2
⇒ AC2 = 52 + 122 = 25 + 144 = 169
⇒ AC = √169 = 13 cm
If AD is bisector of ∠BAC, then
AB/AC = BD/DC
⇒ 5/13 = BD/DC
⇒ DC = 13BD/5
⇒ BD + DC = BC
⇒BD + 13BD/5 = 12
⇒18BD/5 = 12
⇒ BD = 12 × (5/18) = 10/3
In right triangle ABD
AD2 = AB2 + BD2
⇒ AD2 = 52 + (10/3)2 = 25 + 100/3 = 325/9
⇒ AC = √325/9 = 5 √13/3 cm
Q18. The value of  \frac { { sec }^{ 6 }θ-{ tan }^{ 6 }θ-{ 3sec }^{ 2 }θ{ tan }^{ 2 }θ+1 }{ { cos }^{ 4 }θ-{ sin }^{ 4 }θ+{ 2sin }^{ 2 }θ+2 }  
1. \frac { 3}{ 4 }
2. \frac { 2}{ 3 }
3. \frac { 1}{ 2 }
4. 1
Correct answer is Option 2 i.e. \frac { 2}{ 3 }
\frac { { sec }^{ 6 }θ-{ tan }^{ 6 }θ-{ 3sec }^{ 2 }θ{ tan }^{ 2 }θ+1 }{ { cos }^{ 4 }θ-{ sin }^{ 4 }θ+{ 2sin }^{ 2 }θ+2 }
Put θ = 0
⇒ (1 – 0 – 0 + 1)/(1 + 0 + 0 + 2)
⇒ 2/3 (satisfied)
Q.19 Sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E, and sides AD and BC are produced to meet at F. ∠ADC = 75°, and ∠BEC = 52°, then the difference between ∠BAD and ∠AFB is:
1. 21°
2. 31°
3. 22°
4. 23°
Correct answer is Option 2 i.e. 31°
∠ADC = 75° and ∠BEC = 52°
As we know, In a cyclic quadrilateral, the sum of opposite angles are 180°.
∠ADC + ∠ABC = 180°
⇒ ∠ABC = 180° – 75° = 105°
⇒ ∠ABC + ∠CBE = 180° Δ [straight line]
⇒ ∠CBE = 180° – 105° = 75°
In ΔBEC
∠CBE + ∠BEC + ∠ECB = 180°
⇒ ∠ECB = 180° °– 75° – 52° °= 53°
⇒ ∠ECB + ∠BCD = 180°  [straight line]
⇒ ∠BCD = 180° – 53° = 127°
⇒ ∠BAD + ∠BCD = 180°
⇒ ∠BAD = 180° – 127° = 53°
In ΔAFB
∠BAF + ∠ABF + ∠AFB = 180°
⇒ ∠AFB = 180° – 53° – 105° = 22°
∴ Difference between ∠BAD and ∠AFB = 53° – 22° = 31°
Q.20 If 5 sin θ = 4, then then the value of \frac { secθ+4cotθ }{ 4tanθ−5cosθ } is:

1.  \frac { 3 }{ 2 }
2. 1
3. \frac { 5 }{ 4 }
4. 2
Correct answer is Option 4 i.e. 2
5 sin θ = 4
 sin θ = 4/5
Perpendicular = P, Hypotenuse = H and Base = B.
sin θ = P/H = 4/5
P = 4 and H = 5
As we know,
H2 = P2 + B2
⇒ 52 = 42 + B2
⇒ B2 = 25 – 16 = 9
⇒ B = √9 = 3
⇒ cos θ = B/H = 3/5
⇒ sec θ = 1/cos θ = 5/3
tan θ = P/B = 4/3
cot θ = 1/tan θ = 3/4
Now,
\frac { secθ+4cotθ }{ 4tanθ−5cosθ }
= \frac { \frac { 5 }{ 3 } +4\times \frac { 3 }{ 4 } }{ 4\times \frac { 4 }{ 3 }-5 \times \frac { 3 }{ 5 } }
= \frac { \frac { 5 }{ 3 } + 3 }{ \frac { 16 }{ 3 }- 3 }
= \frac { \frac { 14 }{ 3 } }{ \frac { 7 }{ 3 }}
= 2
Q.21 The diagonal of a square A is (a + b) units. What is the area (in square units) of the square drawn on the diagonal of square B whose area is twice the area of A?
1. (a + b) 2
2. 4(a + b)2
3. 8(a + b)2
4. 2(a + b)2
Correct answer is Option 4  i.e. 2(a + b)2
As we know,
Diagonal of square = √2 a
Area of square = a2
Area of square = (diagonal)2/2
Diagonal of square A = (a + b)
Area of square A = (a + b)2/2
Area of square B = 2 × Area of square A
Area of square B = 2 × (a + b)2/2 = (a + b)2
Side of square B = √(a + b)2 = (a + b)
Diagonal of square B = √2 (a + b)
Given,
Side of other square = diagonal of square B = √2 (a + b)
Area of other square = √2 (a + b) × √2 (a + b) = 2 (a + b)2
Other Method:
Let a = b = 1, then
Diagonal of square A = 2
Area of square A = 22/2 = 2
Area of square B = 2 × 2 = 4
Sides of square B = √4 = 2
Diagonal of square B = 2 √2
Given,
Side of other square  = diagonal of square B = 2 √2
Area of other square = 2 √2 × 2 √2 = 8
From option 3
2 (a + b)2
⇒ 2 (1 + 1)2
⇒ 2 × 4
⇒ 8 (satisfied)
Q.22 The given table represents the exports (in Rs. Crores) of four items A, B, C and D over a period of six years. Study the table and answer the question that follows.
The total exports of item D in 2010, 2012 and 2014 is what percentage of the total exports of all the four items in 2011 and 2012?
1. 44.8%
2. 44%
3. 45%
4. 46.2%
 
Correct answer is Option 2 i.e. 44%
The total exports of item D in 2010, 2012 and 2014 = 214 + 247 + 309 = 770
Total exports of all the four items in 2011 and 2012 = 250 + 134 + 244 + 282 + 225 + 138 + 230 + 247 = 1750
∴ Required percentage = 770/1750 × 100 = 44%
Q.23 Pipes A and B can fill a tank in 10 hours and 40 hours, respectively. C is an outlet pipe attached to the tank. If all the three pipes are opened simultaneously, it takes 80 minutes more time than what A and B together take to fill the tank A and B are kept open for 7 hours and then closed and C was opened. C will now empty the tank in:
1. 49 hours
2. 38.5 hours
3. 42 hours
4. 45.5 hours
Corrrect answer is Option 1 i.e. 49 hours

Total work = 40
Time taken to fill the tank A and B together = 40/(1 + 4) = 8 hrs
Let the efficiency of C be x.
If all three pipe A, B and outlet pipe C open, time taken to fill the tank = 8 hr + 80 min = 9 + 1/3 = 28/3 hrs
Efficiency of A,  B and C = 40/(28/3) = 30/7
According to the question
4 + 1 + x = 30/7
⇒ x = 30/7 – 5
⇒ x = (–5/7)
Work done by A and B in 7 hrs = (4 + 1) × 7 = 35
∴ Pipe C empty the tank filled by A and B in 7 hrs = 35/(5/7) = 49 hrs
Q.24 The compound interest on a certain sum at 16\frac { 2 }{ 3}% p. a for 3 years is Rs. 6,350 What will be the simple interest on the same at the same rate for 5\frac { 2 }{ 3}  years?
 
1. Rs. 10,200
2. Rs. 11,400
3. Rs. 4,620
4. Rs. 9, 600
Correct answer is Option 1 i.e. Rs. 10,200
Detailed Method:
Rate = 16\frac { 2 }{ 3}% \frac { 50 }{ 3}%
Time = 3 years
CI = 6350
As we know,
CI = P [(1 + r/100)3 – 1]
6350 = P [{1 + (50/3 × 100)3} – 1]
⇒ 6350 = P [(1 + 1/6)3 – 1]
⇒ 6350 = P [(7/6)3 – 1]
⇒ 6350 = P [343/216 – 1]
⇒ 6350 = P [127/216]
⇒ P = 6350 × (216/127)
⇒ P = 10800
Now,
P = 10800,  r= 16\frac { 2 }{ 3}% \frac { 50 }{ 3}%
and T5\frac { 2 }{ 3} years = \frac { 17 }{ 3} years
As we know,
SI = Prt/100
SI = (10800 × 50 × 17)/(3 × 3 × 100)
SI = 10200
Short trick:
Rate = 16\frac { 2 }{ 3}% = \frac { 1 }{ 6}
and time = 3 years and CI = 6350
Ratio of P to A = 63 : 73 = 216 : 343
⇒ 343 – 216 = 127
⇒ 127 unit = 6350
⇒ 1 unit = 50
⇒ 216 unit = 10,800
Now,
P = 10800,
r= 16\frac { 2 }{ 3}% \frac { 50 }{ 3}%
and T5\frac { 2 }{ 3} years = \frac { 17 }{ 3} years
Interest rate for 1 year = 50/3 %
Interest rate for 17/3 year = 850/9 %
Simple interest = 10800 × (850/900) = 10200
Hint:
As given, time = 17/3 years
So, we can say answer divisible by 17. Only option 2 is divisible by 17.
Q.25 \frac { 3\frac { 2 }{ 3 } \div \frac { 11 }{ 30 } of\frac { 2 }{ 3 } -\frac { 1 }{ 4 } of2\frac { 1 }{ 2 } \div \frac { 3 }{ 5 } \times 4\frac { 4 }{ 5 } }{ \frac { 2 }{ 5 } of 7\frac { 1 }{ 2 } \div \frac { 3 }{ 4 }- \frac { 3 }{ 4 } \times 1\frac { 1 }{ 2 } \div 2\frac { 1 }{ 4 } } is:

1. 2\frac { 6 }{ 7 }
2. 2\frac {2 }{ 9}
3. 3\frac { 4 }{ 7 }
4. \frac { 10 }{ 21 }

Correct answer is Option 1  i.e. 2\frac { 6 }{ 7 }

\frac { 3\frac { 2 }{ 3 } \div \frac { 11 }{ 30 } of\frac { 2 }{ 3 } -\frac { 1 }{ 4 } of2\frac { 1 }{ 2 } \div \frac { 3 }{ 5 } \times 4\frac { 4 }{ 5 } }{ \frac { 2 }{ 5 } of 7\frac { 1 }{ 2 } \div \frac { 3 }{ 4 }- \frac { 3 }{ 4 } \times 1\frac { 1 }{ 2 } \div 2\frac { 1 }{ 4 } }

= \frac { \frac { 11 }{ 3 } \div \frac { 11 }{ 30 } \times \frac { 2 }{ 3 } -\frac { 1 }{ 4 } \times \frac { 5 }{ 2 } \div \frac { 3 }{ 5 } \times \frac { 24 }{ 5 } }{ \frac { 2 }{ 5 } \times \frac { 15 }{ 2 } \div \frac { 3 }{ 4 } -\frac { 3 }{ 4 } \times \frac { 3 }{ 2 } \div \frac { 9 }{ 4 } }

= \frac { \frac { 11 }{ 3 } \div \frac { 11 }{ 45 } -\frac { 5 }{ 8 } \div \frac { 3 }{ 5 } \times \frac { 24 }{ 5 } }{ 3\div \frac { 3 }{ 4 } -\frac { 3 }{ 4 } \times \frac { 3 }{ 2 } \times \frac { 4 }{ 9 } }

= \frac { \frac { 11 }{ 3 } \div \frac { 45 }{ 11 } -\frac { 5 }{ 8 } \times \frac { 5 }{ 3 } \times \frac { 24 }{ 5 } }{ 3\times \frac { 4 }{ 3 } -\frac { 1 }{ 2 } }

= \frac { 15-5 }{ 4-\frac { 1 }{ 2 } }

= \frac { 10 }{ \frac { 7 }{ 2 } }

= \frac { 20 }{ 7 }

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
http://edumo.in

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