SSC CGL 2019 EXAM PAPER : Held on 04-March-2020 Shift-1 (Quantitative Aptitude)

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SSC CGL 2019 EXAM PAPER : Held on 04-March-2020 Shift-1 (Quantative Aptitude)

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Q.1 ∆ABC, ∠A = 90°, M is the midpoint of BC and D is a point on BC such that AD⊥ BC. If AB = 7 cm and AC = 24 cm, then AD : AM is equal to:

1. 32 : 43
2. 24 : 25
3. 336 : 625
4. 168 : 275
Correct answer is Option 3 i.e.  336 : 625
In right angled ∆ABC, 90° at A
BC2 = AB2 + AC2 = 72 + 242 = 49 + 576 = 625
⇒ BC = √625 = 25
If, M is the midpoint of BC, then
AM = MC = BM = 25/2
As we know,
AD × BC = AB × AC
⇒ AD × 25 = 7 × 24
⇒ AD = (7 × 24)/25
∴ Ratio of AD : AM = (7 × 24)/25 : 25/2 = 336 : 625
Q.2 On simplification,  \frac { { x }^{ 3 }-{ y }^{ 3 } }{ { x[(x+y) }^{ 2 }-3xy] } \div \frac { { y[(x−y) }^{ 2 }+3xy] }{ { x }^{ 3 }+{ y }^{ 3 } } \times \frac { { (x+y) }^{ 2 }-{ (x−y) }^{ 2 } }{ {x}^{2}-{y}^{2} } is equal to :
1. 4
2. 1 
3. \frac {1}{2}
4. \frac {1}{4}

The correct answer is option 1 i.e  4

As we know,

(x3 + y3) = (x + y) (x2 + y2 – xy)

(x3 – y3) = (x – y) (x2 + y2 + xy)

(x + y)2 – (x – y)2 = 4xy

[(x – y)2 + 3xy] = [x2 + y2 – 2xy + 3xy] = [x2 + y2 + xy]

[(x + y)2 – 3xy] = [x2 + y2 + 2xy – 3xy] = [x2 + y2 – xy]

Now,

\frac { { x }^{ 3 }-{ y }^{ 3 } }{ { x[(x+y) }^{ 2 }+3xy] } \div \frac { { y[(x−y) }^{ 2 }-3xy] }{ { x }^{ 3 }+{ y }^{ 3 } } \times \frac { { (x+y) }^{ 2 }-{ (x−y) }^{ 2 } }{ { x }^{ 2 }-{ y }^{ 2 } }

=\frac { { (x−y)(x }^{ 2 }+{ y }^{ 2 }+xy) }{ { x[(x }^{ 2 }+{ y }^{ 2 }+xy)] } \div \frac { { y[(x }^{ 2 }+{ y }^{ 2 }-xy)] }{ { (x+y)(x }^{ 2 }+{ y }^{ 2 }-xy) } \times \frac { 4xy }{ (x−y)(x+y) }

=\frac { { (x−y)(x }^{ 2 }+{ y }^{ 2 }+xy) }{ { x[(x }^{ 2 }+{ y }^{ 2 }+xy)] } \times \frac { { (x+y)(x }^{ 2 }+{ y }^{ 2 }-xy) }{ { y(x }^{ 2 }+{ y }^{ 2 }-xy) } \times \frac { 4xy }{ (x−y)(x+y) }

=4

Q.3 If 5x+\frac { 1 }{ 3x }=4 then what is the value of { 9x }^{ 2 }+\frac { 1 }{ { 25x }^{ 2 } } ?
1. \frac {174 }{125 }
2. \frac {119 }{25 }
3. \frac {144 }{125 }
4. \frac {114 }{25 }
Correct answer is option 4 i.e.  \frac {114 } {25 }
5x + 1/3x = 4
Multiply by 3/5
3x + 1/5x = 12/5
Squaring on both sides
(3x + 1/5x)2 = (12/5)2
⇒ 9x2 + 1/25x2 + 2 × 3x × (1/5x) = 144/25
⇒ 9x2 + 1/25x2 = 144/25 – 6/5
⇒ 9x2 + 1/25x2 = (144 – 30)/25 = 114/25
Q.4 The given table represents the number of computers sold by four dealers A, B, C and D during the first six months of 2016. Study the table carefully and answer the question that follows.
1. 25.3
2. 21.2
3. 24.4
4. 17.5
Correct answer is Option 2 i.e. 21.2
Total computers sold by dealer A from February to June = 94 + 85 + 108 + 98 + 95 = 480
Total number of computers sold by all the dealers in June = 95 + 108 + 102 + 91 = 396
∴ Required percentage [(480 – 396)/396] × 100 = 21.2
Q.5 The average of twelve numbers is 45.5. The average of the first four numbers is 41.5 and that of the next five numbers is 48. The 10th number is 4 more than the 11th number and 9 more than the 12th number. What is the average of the 10th and 12th numbers?
1. 47
2. 47.8
3. 46
4. 46.5
Correct answer is option 4 i.e. 46.5
Average of 12 numbers in = 45.5
Sum of 12 numbers = 45.5 × 12 = 546
Average of first 4 numbers = 41.5
Sum of first 4 numbers = 41.5 × 4 = 166
Average of next 5 numbers = 48
Sum of next 5 numbers = 48 × 5 = 240
Sum of 10th, 11th and 12th number = 546 – 166 – 240 = 140
Let the 10th number be x
11th number = x – 4
12th number = x – 9
According to the question
x + x – 4 + x – 9 = 140
⇒ 3x – 13 = 140
⇒ 3x = 140 + 13
⇒ 3x = 153
∴ x = 153/3 = 51
Average of 10th and 12th number = (51 + 51 – 9)/2 = 93/2 = 46.5
Q.6 The value of  \frac { tan30°cosec60°+tan60°sec30° }{ { sin }^{ 2 }30°+{ 4cot }^{ 2 }45°-{ sec }^{ 2} 60°} is:

1. \frac { 2 }{ 3 }
2. \frac { 32 }{ 99 }
3. \frac { 8 }{ 3 }
4. \frac { 32 }{ 3 }
Correct answer is option 4 i.e. \frac { 32 }{ 3 }
\frac { tan30°cosec60°+tan60°sec30° }{ { sin }^{ 2 }30°+{ 4cot }^{ 2 }45°-{ sec }^{ 2} 60°}
=\frac { \frac { 1 }{ \sqrt { 3 } } \times \frac { 2 }{ \sqrt { 3 } } +\sqrt { 3 } \times \frac { 2 }{ \sqrt { 3 } } }{ \frac { 1 }{ { 2 }^{ 2 } } +4\times 1-{ 2 }^{ 2 } } 
=\frac { \frac { 2 }{ 3 } +2 }{ \frac { 1 }{ 4 } +4-4 }
=\frac { 8 }{ 3 } \times 4
\frac { 32 }{ 3 }
Q.7 Sonu saves 15% of her income. If her income increases by 20% and she still saves the same amount as before, then what is the percentage increase in her expenditure? (correct to one decimal place)
1. 23.5
2. 22.8
3. 23.8
4. 24.2
Correct answer is option 1 i.e. 23.5
Let the income of Sonu be Rs. 100
Savings of Sonu = 100 × (15/100) = Rs. 15
Expenditure of Sonu = 100 – 15 = Rs. 85
New income of Sonu = 100 × (120/100) = Rs. 120
If savings amount same, then new expenditure = 120 – 15 = 105
Her expenditure increased by = 105 – 85 = 20
Her expenditure increased by = 20/85 × 100 ≈ 23.5%
Q.8 When 732 is divided by a positive integer x, the remainder is 12. How many values of x are there?
1. 19
2. 18
3. 20
4. 16
Correct answer is option 3 i.e. 20
When 732 is divided by a positive integer x, the remainder is 12. So,
732 – 12 = 720
So 720 will be completely divisible by x
720 = 24 × 32 × 51
Total number of factors of 720 = (4 + 1) × (2 + 1) × (1 + 1) = 5 × 3 × 2 = 30
Number of factors 12 or less than 12 are (10) = 1, 2, 3, 4, 5, 6, 8, 9, 10, 12
So x cannot have these values because x is greater than 12
So, we can say that there are 30 – 10 = 20 possible values of x
Q.9 The circumference of the base of a conical tent is 66 m. If the height of the tent is 36 m, what is the area (in m2) of the canvas used in making the tent? (Take π = 22/7)

1. 1237.5
2. 1171.5
3. 1254
4. 1155
 
Correct answer is option 1 i.e 1237.5
Height of the conical tent h = 36 m
As we know
Curved surface area of conical tent = πrl
Circumference of the base of a conical tent = 2 πr
2 πr = 66
⇒ 2 × (22/7) × r = 66
⇒ r = 66 × (7/22) × (1/2)
⇒ r = 21/2
As we know,
l2 = r2 + h2
⇒ l2 = (21/2)2 + 362
⇒ l2 = 441/4 + 1296
⇒ l2 = (441 + 5184)/4 = 5625/4
⇒ l = √5625/4
⇒ l = 75/2
∴ Curved surface area of conical tent = πrl = (22/7) × (21/2) × (75/2) = 1237.5 m2
Q10. If a + b + c = 11, ab + bc + ca = 3 and abc = -135, then what is the value of a3 + b3 + c3?

 

1. 827
2. 929
3. 823
4. 925
Correct answer is Option 1. i.e. 827 
a + b + c = 11, ab + bc + ca = 3 and abc = -135
As we know,
a3 + b3 + c3 – 3abc = (a + b + c) [(a + b + c)2 – 3(ab + bc + ca)]
⇒ a3 + b3 + c3 – 3 × (-135) = 11 [112 – 3 × 3]
⇒ a3 + b3 + c3 + 405 = 11 × [121 – 9] = 11 × 112
⇒ a3 + b3 + c3 = 1232 – 405 = 827
Q.11 ∆ABC, AB = AC. A circle drawn through B touches AC at D and intersects AB at P. If D is the midpoint of AC and AP = 2.5 cm, then AB is equal to:

1. 9 cm
2. 10 cm
3. 7.5 cm
4. 12.5 cm
correct answer is option 2 i.e. 10cm
AB = AC
AC = 2 AD
As we know,
AD2 = AP × AB
⇒ AD2 = 2.5 × AC
⇒ AD2 = 2.5 × 2AD
AD = 5 cm
⇒ AC = 2 × 5 = 10 cm
∴ AB = AC = 10 cm
Q.12 If 5sin2θ + 14 cosθ = 13, 0° < θ < 90°, then what is the value of  \frac { secθ+cotθ }{ cosecθ+tanθ } ?
1. \frac { 31}{29 }
2. \frac { 32}{27 }
3. \frac {21 }{28 }
4. \frac {9 }{8 }
Correct answer is Option 1 i.e.  \frac { 31}{29 }
5 sin2 θ + 14 cos θ = 13
⇒ 5 (1 – cos2 θ) + 14 cos θ = 13
⇒ 5 – 5 cos2 θ + 14 cos θ = 13
⇒ 5 cos2 θ – 14 cos θ + 8 = 0
⇒ 5 cos2 θ – 10 cos θ – 4 cos θ + 8 = 0
⇒ 5 cos θ (cos θ – 2) – 4 (cos θ – 2) = 0
⇒ (5 cos θ – 4) (cos θ – 2) = 0
Taking,
(5 cos θ – 4) = 0 (as cosθ can’t be 2)
cosθ = 4/5
Perpendicular = P, Hypotenuse = H and Base = B.
cos θ = B/H = 4/5
H = 5 and B = 4
As we know,
H2 = P2 + B2
⇒ 52 = P2 + 42
⇒ P2 = 25 – 16 = 9
⇒ P = √9 = 3
sec θ = H/B = 5/4
cot θ = B/P = 4/3
cosec θ = H/P = 5/3
tan θ = P/B = 3/4
Now,
(sec θ + cot θ)/(cosec θ + tan θ)
⇒ (5/4 + 4/3)/(5/3 + 3/4)
⇒ [(15 + 16)/12]/[(20 + 9)/12]
31/29
Q.13 The value of \frac { 7−[4+3(2−2×2+5)−8]÷5 }{ 2÷2of(4+4÷4of4) } is :
1. 8\frac { 1}{ 2 }
2. 24
3. 25\frac { 1}{ 2 }
4. 26

Correct Answer is option 3 i.e. 25\frac { 1}{ 2 }

\frac { 7−[4+3(2−2×2+5)−8]÷5 }{ 2÷2of(4+4÷4of4) }

= \frac { 7−[4+3(2−4+5)−8]÷5 }{ 2÷2of(4+4÷16) }

=\frac { 7−[4+3×3−8]÷5 }{ 2÷2of(4+\frac { 1 }{ 4 } ) }

= \frac { 7−[4+9−8]÷5 }{ 2÷2×\frac { 17 }{ 4 } ) }

= \frac { 7−5÷5 }{ 2÷\frac { 17 }{ 2 } }

= \frac { 7−1}{ 2\times \frac { 2 }{ 17 } }

= 6\times \frac { 17}{ 4 }

= \frac { 51}{ 2 }

= 25\frac { 1}{ 2 }

Q.14 If 2x + 1, x + 2, 2 and 5 are in proportion, then what is the mean proportional between 3.5 (1 – x) and 8 (1 + x)?

1. 5.5
2. 5.25
3. 4.25
4. 4.5
Correct answer is Option 2 i.e. 5.25
If 2x + 1, x + 2, 2 and 5 are in proportion, then
2x + 1 : x + 2 :: 2 : 5
⇒ (2x + 1)/(x + 2) = 2/5
⇒ 5 (2x + 1) = 2 (x + 2)
⇒ 10x + 5 = 2x + 4
⇒ 10x – 2x = 4 – 5
⇒ 8x = (-1)
⇒ x = -1/8
Let mean proportional between 3.5 (1 – x) and 8 (1 + x) be M, then
3.5 (1 – x) : M :: M : 8 (1 + x)
⇒ [3.5(1 + 1/8)]/M = M/[8 (1 – 1/8)]
⇒ M2 = 3.5 × 9/8 × 8 × 7/8
⇒ M2 = (7 × 7 × 3 × 3)/(4 × 4)
⇒ M = √(7 × 7 × 3 × 3)/(4 × 4)
⇒ M = (7 × 3)/4
⇒ M = 21/4
∴ M = 5.25
Q.15 ABCD is a cyclic quadrilateral in which AB = 16.5 cm, BC = x cm, CD = 11 cm, AD = 19.8 cm, BD is bisected by AC at O. What is the value of x?
1. 12.8 cm
2. 12.4 cm
3. 13.2 cm
4. 13.8 cm
Correct answer is Option 3 i.e. 13.2 cm
From the following figure
As we know,
ΔAOB∼ΔDOC
AB/DC = OB/OC
⇒ OB/OC = 16.5/11
Given, OB = OD
OD/OC = 16.5/11     …1)
As we know,
ΔABO∼ΔDCO
OD/OC = AD/BC
⇒ OD/OC = 19.8/x     …2)
From equation (1) and equation (2)
19.8/x = 16.5/11
⇒ x = (11 × 19.8)/16.5
⇒ x = 13.2 cm
Q.16 In ∆ABC, ∠B = 68° and ∠C = 32°. Sides AB and AC are produced to points D and E respectively. The bisectors of ∠DBC and ∠BCE meet at F. what is the measure of ∠BFC?
1. 39°
2. 65°
3. 55°
4. 50°
Correct answer is Option 4 i.e. 50°
In ∆ABC
∠ABC = 68° and ∠ACB = 32°
As we know,
∠ABC + ∠DBC = 180°
⇒ ∠DBC = 180° – 68°= 112°
⇒ ∠CBF = 112°/2 = 56°
Similarly,
∠ACB + ∠BCE = 180°
⇒ ∠BCE = 180° – 32° = 148°
⇒ ∠BCF = 148°/2 = 74°
In ∆BFC
∠BFC + ∠CBF + ∠ BCF = 180°
⇒ ∠BFC = 180° – 56° – 74° = 50°
Short Trick:
∠B = 68° and ∠C° = 32°
∠A = 180° – 68° – 32° = 80°
Now, ∠BFC = 90° – ∠A/2 = 90° – 80°/2 = 50°
Q.17 A trader allows a discount of 18% on the marked price of an article. How much percentage above the cost price must he mark it so as to get a profit of 6.6%?
1.  30
2.  24
3.  28
4.  25
Correct answer is Option 1 i.e.  30
Short trick:
Let the marked price be decided to be x% above the cost price, then
6.6 = x – 18 – 18x/100
⇒ 6.6 + 18 = x – 0.18x
⇒ 0.82x = 24.6
⇒ x = 24.6/0.82 = 30
Detailed Method:
Let the Cost price of the article be Rs. 100, then
SP of the article = 100 × (106.6) = Rs. 106.6
MP of the article = 106.6 × (100/82) = 130
MP is more than CP by = 130 – 100 = 30
∴ MP is more than CP by (in %) = 30/100 × 100 = 30%
Q18. The given table represents the number of computers sold by four dealers A, B, C and D during the first six months of 2016. Study the table carefully and answer the question that follows.
The total number of computers sold by dealer B in April, May and June is what percentage of the total number of computers sold by all the dealers in February and April?
1. 48\frac { 5}{ 7 }
2. 43\frac { 6}{ 7 }
3. 38\frac { 3}{ 8 }
4. 50\frac { 7}{ 8 }
Correct answer is Option 3 i.e. 38\frac { 3}{ 8 }  
Total number of computer sold by dealer B in April, May and June = 97 + 102 + 108 = 307
Total number of computer sold by all the dealers in February and April = 94 + 96 + 104 + 106 + 108 + 97 + 99 + 96 = 800
∴ Required percentage = 307/800 × 100 =
38\frac { 3}{ 8 }  
Q.19 A boat can go 3 km upstream and 5 km downstream in 55 minutes. It can also go 4 km upstream and 9 km downstream in 1 hour 25 minutes. In how much time (in hours) will it go 43.2 km downstream?
1. 3.6
2. 4.4
3. 5.4
4. 4.8
Correct answer is Option 1 i.e. 3.6
Let the speed of boat and current be x km/hr and y km/hr respectively.
Downstream speed = (x + y) km/hr
Upstream speed = (x – y) km/hr
According to the question
3/(x – y) + 5/(x + y) = 55/60 = 11/12
4/(x – y) + 9/(x + y) = 1 + 25/60 = 17/12
Let 1/(x + y) = a and 1/(x – y) = b
3b + 5a = 11/12     …1)
4b + 9a = 17/12     …2)
Multiply by 4 in equation (1) and multiply by 3 in equation (2)
12b + 20a = 11/3     …3)
12b + 27a = 17/4     …4)
Subtract equation (3) from equation (4)
7a = 17/4 – 11/4
⇒ 7a = 7/12
⇒ a = 1/12
⇒ x + y = 12 km/hr
Speed of downstream = 12 km/hr
∴ Time taken to cover 43.2 km distance in downstream = 43.2/12 = 3.6 hr
Q.20 What is the compound interest on a sum of Rs. 12,000 for 2\frac { 5}{ 8 }  years at 8% p.a. when the interest is compounded annually (nearest to a rupee)?

1. Rs. 2,642
2. Rs. 2,697
3. Rs. 2,712
4. Rs. 2,654
Correct answer is Option 2 i.e. Rs. 2,697
Let the five consecutive even numbers are 2, 4, 6, 8, 10.
Average of consecutive even numbers = (2 + 4 + 6 + 8 + 10)/5 = 30/5 = 6
Hint: As we know, Average of any AP is the median so, the average = 6
So, M = 6
If the next five even numbers are also included, then the numbers are
2, 4, 6, 8, 10, 12, 14, 16, 18, 20
Average of 10 consecutive numbers = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20)/10 = 110/10 = 11
Hint: As we know, Average is the median so, the average = 11
We can write 11 = 6 + 5 or M + 5

Q.21 If sec θ – tan θ = x/y, (0 < x < y) and 0°< θ < 90°, then sinθ is equal to:

1. \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { y }^{ 2 }-{ x }^{ 2 } }

2. \frac { { x }^{ 2 }+{ y }^{ 2 } }{2xy }

3. \frac { { y }^{ 2 }-{ x }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } }

4. \frac {2xy }{ { x }^{ 2 }+{ y }^{ 2 } }

Correct answer is Option 3  i.e. \frac { { y }^{ 2 }-{ x }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } }

sec θ – tan θ = x/y     …1)

As we know,

sec2 θ – tan2 θ = 1

⇒ (sec θ + tan θ) (sec θ – tan θ) = 1

⇒ (sec θ + tan θ) x/y = 1

⇒ (sec θ + tan θ) = y/x     …2)

Add equation (1) and equation (2)

2 sec θ = x/y + y/x

⇒ sec θ = (x2 + y2)/2xy

⇒ cos θ = 2xy/(x2 + y2) [∵ sec θ = 1/cosθ]

As we know,

sin2 θ + cos2 θ = 1

⇒ sin2 θ + [2xy/(x2 + y2)]2 = 1

⇒ sinθ = 1 – 4x2y2/(x2 + y2)2

⇒ sin2 θ = [(x2 + y2)2 – 4x2y2]/(x2 + y2)2

⇒ sin2 θ = [x4 + y4 + 2x2y2 – 4x2y2]/(x2 + y2)2

⇒ sin2 θ = [x4 + y4 – 2x2y2]/(x2 + y2)2

⇒ sin2 θ = [(y2 – x2)/(x2 + y2)]2 [∵ x < y]

∴ sin θ = (y2 – x2)/(x2 + y2)

Q.22 A person buys 5 tables and 9 chairs for Rs. 15,400. He sells the tables at 10% profit and chairs at 20% profit. If his total profit on selling all the tables and chairs is Rs. 2,080 what is the cost price of 3 chairs?
1. Rs. 1,800
2. Rs. 1,740
3. Rs. 1,860
4. Rs. 1,890
 
Correct answer is Option 2 i.e. Rs. 1,800
10% of 15400 = 1540
20% of 15400 = 3080
Using allegation method
1540 3080
2080
1000 : 540
Ratio of cost price of 5 tables and 9 chairs = 1000x : 540x
1000x + 540x = 1540x
⇒ 1540x = 15400
⇒ x = 15400/1540
⇒ x = 10
Cost price of 9 chairs = 540 × 10 = 5400
⇒ Cost price of 1 chair = 5400/9 = 600
∴ Cost price of 3 chairs = 3 × 600 = 1800
Detailed Method:
Let the cost price of one table and one chair be Rs. x and Rs. y respectively.
5x + 9y = 15400     …1)
According to the question
5x × (10/100) + 9y × (20/100) = 2080
⇒ 5x/10 + 18y/10 = 2080
⇒ 5x + 18y = 20800     …2)
Subtract equation (1) from equation (2).
9y = 5400
⇒ 3y = 1800
∴ Cost of 3 chairs are Rs. 1,800.
Q.23 The given table represents the number of computers sold by four dealers A, B, C and D during the first six months of 2016. Study the table carefully and answer the question that follows.
The number of months, in which the number of computers sold by dealer B was less than the average number of computers sold by dealer C over six months, is:
1. 4
2. 3
3. 5
4. 2
Corrrect answer is Option 1 i.e. 4
Total number of computer sold by dealer C over six months = 95 + 104 + 100 + 99 + 100 + 102 = 600
Average computer sold by dealer C over six months = 600/6 = 100
There are four months January, February, March and April in which the number of computers sold by dealer B was less than the average number of computers sold by dealer C in six months.
Q.24 The given table represents the number of computers sold by four dealers A, B, C and D during the first six months of 2016. Study the table carefully and answer the question that follows.
What is the ratio of the total number of computers sold by dealer A in February, April and May to the total number of computers sold by dealer D in March, May and June?
1. 10 : 9
2.  6 : 5
3. 15 : 13
4. 20 : 27
Correct answer is Option 1 i.e. 10 : 9
Total number of computer sold by seller A in February, April and May = 94 + 108 + 98 = 300
Total number of computer sold by seller D in March, May and June = 90 + 89 + 91 = 270
∴ Required ratio = 300 : 270 = 10 : 9
Q.25 A can complete a certain work in 30 days. B is 25% more efficient than A and C is 20% more efficient than B. They all worked together for 3 days. B alone will complete the remaining work in:

1. 18 days
2. 12 days
3. 20 days
4. 15 days
Correct answer is Option 4  i.e. 15 days.
A can complete work in = 30 days
Efficiency of A (A’s 1 day’s work) = 1/30
Efficiency of B (B’s 1 day’s work) = 1/30 × (125/100) = 1/24
Efficiency of C (C’s 1 day’s work) = 1/24 × (120/100) = 1/20
Let the remaining work be complete by B in x days.
According to the question
3 (1/30 + 1/24 + 1/20) + x/24 = 1
⇒ 3 [(4 + 5 + 6)/120] + x/24 = 1
⇒ 15/40 + x/24 = 1
⇒ x/24 = 1 – 3/8 ×
⇒ x/24 = (8 – 3)/8
⇒ x/24 = 5/8
⇒ x = 5/8 × 24 = 15
Short Trick:
Let Efficiency of A = 4
Efficiency of B = 4 × 5/4 = 5
Efficiency of C = 5 × 6/5 = 6
Total work = 30 × 4 = 120
Work done by A, B and C in 3 days = (4 + 5 + 6) × 3 = 45
Remaining work = 120 – 45 = 75
∴ B completes the remaining work in = 75/5 = 15 days

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

Manish Aggarwal
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