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Solutions

S1. Ans.(b)

Sol. Total students of class V in 2012 = 140

70 × 100 = 200

Failed students of class V in 2013 = 140 × 100

175 = 80

∴ Required ratio = 30

100 × 200

80

100 ×

80

20 × 100

=3

16

S2. Ans.(c)

Sol. Total strength in 2013 = 150 + 150 = 300

∴ Number of passed students = 60

100 × 300 = 180

S3. Ans.(a)

Sol. Failed girls in 2014 = 7 × 5 = 35

Failed boys in 2014 = 35 × 10

7 = 50

∴ Total students in 2014 = 85 × 100

34 = 250

Total students in 2010 = 250 × 4

5 = 200

Number of passed girls in 2010 = 84 × 100

120 = 70

∴ Number of passed boys in 2010 = 65

100 × 200 – 70

= 60

S4. Ans.(d)

Sol. Passed students in class III = 105

42 × 58 = 145

Failed students in class IV = 198

88 × 12 = 27

Passed girls in class III = 145 − 17

2 = 64

Failed boys in class IV = 27 + 15

2 = 21

∴ Required difference = 43

S5. Ans.(c)

Sol. Average = 1

5 (78 + 82 + 88 + 60 + 55)

= 1

5 × 363

= 72.6%


S6. Ans.(c)

Sol. Let number of boys and girls are 2x and 3x respectively

Increase in number of boys =2x×120

100=

12x

5

New ratio = 4 : 5

Now 20 girls more added then ratio become 3 : 5

Equal boys proportion

(12 + 20) × 4 = 128

S7. Ans.(c)

Sol. Let the initial investments of Ramesh, Rajan, Ritesh be Rs. 3x, 5x and 7x respectively. Then,

(3x – 45600) : 5x : (7x + 337600) = 24 : 59 : 167

⇒3x − 45600

5x=

24

59⇒ x = 47200.

∴Ramesh initially invested Rs. (47200 × 3) = Rs. 141600

S8. Ans.(a)

Sol. Average age of class = 16

Average age of boys =21

Average age of girls = 12

Now by allegation

Required ratio = 4 : 5

S9. Ans.(d)

Sol. F: S=x: (50 − x)

Eight years ago, x − 8: 42 − x

From question ->(x − 8)(42 − x) = 2(x − 8)

x = 40, So father’s age = 40, son = 10


S10. Ans.(c)

Sol. Let the amount of pure copper = x kg.

Pure copper + copper in 1st alloy + copper in 2nd alloy

= Copper in 3rd alloy

⇒ x + 5

7×14 +

2

5×10 =

2

3(14 + 10 + x)

⇒ 14 + x = 2

3 (24 + x)

⇒ x = 6 kg.

S11. Ans.(b)

Sol. D

4−

D

5=

15

60, D (

5−4

20) =

15

60, D =

20×15

60= 5km

S12. Ans.(c)

Sol. From the given information there could be many solution for given question.

S13. Ans.(a)

Sol. Quadrilateral which have all side equal and none of the angle of right angle, then it is a

Rhombus.

Area of Rhombus = 1

2d1×d2

Also, Area of Rhombus = base × height

= 6.5 × 10

= 65

65 = 1

2×d1×d2

d2 =65 × 2

26= 5cmv

S14. Ans.(d)

Sol. Let initial expenditures an clothes, electricity and fuel be Rs. 12x, Rs. 17x and Rs 3x respectively.

Total expenditure = 12x +17x+3x=Rs. 32x

After increase

Expenditure an clothes = 120

100×12x = Rs. 14.4x

Expenditure an electricity = 130

100×17x = Rs. 22.1x

Expenditure an fuel = 150

100×3x = 4.5x


Total expenditure = 14.4x + 22.1x + 4.5x

= 41x

Increase = 9x

Percentage Increase = 9x

32x×100 = 28

1

8%

S15. Ans.(e)

Sol. Reqd. Probability =4c1+3c1

19c1=

7

19

S16. Ans.(a)

Sol. Total population of females and transgenders in village P in 2000 = 75% of 2400 = 1800

∴Number of females in village P in 2000 =3

10×1800 = 540

Females in 2001 in village P = 540×120

100= 648

∴ Total males & transgenders in 2001 in village P = 2400 – 648 = 1752

S17. Ans.(c)

Sol. Percentage transgenders in village R in year 2000 = 30%

∴ Total population of village R in 2000 =180

30×100 = 600

∴males in village R in 2000 = 600×50% =300

Males in village S in 2000 =84

100×800×

1

3= 224

∴Required difference = 300 – 224 = 76

S18. Ans.(b)

Sol. Total population of village Q and Village R in 2000 = 2400×125

100= 3000

∴ Total population of village Q in 2000 =2

5×3000 = 1200 and

total population of village R in 2000 =3

5×3000 = 1800

∴ Required ratio =

40

100×1200

30

100×1800

=4×2

3×3=

8

9= 8 ∶ 9

S19. Ans.(d)

Sol. Cannot be determined

S20. Ans.(e)

Sol. Let the population of R=5x

And the population of T = 4x

Required percentage =(4x)×

40

100− (5x)×

30

100

(5x)×30

100

×100 =(1.6 − 1.5)x

(1.5)x×100 =

0.1×100

1.5= 6.667%


S21. Ans.(b)

Sol. I. x² = 144

⇒ x = ±12

II. y² - 24y + 144 = 0

⇒ (y – 12)² = 0

⇒ y – 12 = 0

⇒ y = 12

So, x ≤ y

S22. Ans.(b)

Sol. I. 2x² - 9x + 10 = 0

⇒ (x – 2) (2x - 5) = 0

⇒ x = 2 or5

2

II. 2y² - 13y + 20 = 0

⇒ (y - 4) (2y - 5) = 0

⇒ y = 4 or5

2

∴ y ≥ x

S23. Ans.(a)

Sol. I. 2x² + 15x + 27 = 0

⇒ (2x + 9) (x + 3) = 0

⇒ x =−9

2 or − 3

II. 2y² + 7y + 6 = 0

⇒ (2y + 3) (y + 2) = 0

⇒ y = −3

2 or − 2

∴ x < y

S24. Ans.(e)

Sol. I. 3x² - 13x + 12 = 0

⇒ (3x – 4) (x - 3)= 0

⇒ x =4

3 or 3

II. 3y² - 13y + 14 = 0

⇒ (3y - 7) (y - 2) = 0

⇒ y =7

3or 2

So, no relation exist between x and y


S25. Ans.(d)

Sol. I. 5x² + 8x + 3 = 0

⇒ (5x + 3) (x + 1) = 0

⇒ x =−3

5 or − 1

II. 3y² + 7y+ 4= 0

⇒ (y + 1) (3y + 4) = 0

⇒ y = −1 or−4

3

So, x ≥ y

S26. Ans.(e)

Sol. Let work done by 15 men in 9 days = W2

⇒18×30

1=

15×9

W2

⇒ W2 =15×9

18×30=

1

4

∴ Remaining work = 1 −1

4=

3

4

16 women can complete the same project in 36 days

⇒16×36

1=

18×D23

4

18×D2 =3

4×16×36

D2 = 24 days

S27. Ans.(d)

Sol. Suppose, the initial quantity of mixture in the vessel = 4x + 6x + 5x = 15x litre

The quantity of grape juice in 15 litre of mixture = 4 litre

The quantity of pineapple juice in 15 litre of mixture = 6 litre

The quantity of banana shake in 15 litre of mixture = 5 litre

According to question, (6x − 6 + 2) − (4x − 4 + 8) = 10 ⇒ 2x = 10 + 8 ⇒ x = 9 litre

∴ The initial quantity of mixture = 15 × 9 = 135 litre

S28. Ans.(a)

Sol. Let the distance between A and B = d km

According to question, d

18−

d

24= 1

4d−3d

72= 1

d = 72 km

Time taken by Sara with the speed of 18 km/h = 72

18= 4 hour

∴ Required speed = 72

4−2=

72

2= 36 km/h


S29. Ans.(e)

Sol. Probability that first ball is red = 6

22=

3

11

Probability that second ball is yellow = 11

21

∴ Required probability = 3

11×

11

21=

1

7

S30. Ans.(a)

Sol. Let radius of cylinder = r metre

Total surface area of cylinder = 2πr(r + h)

1672 = 2×22

7×r(r + 19 − r)

r = 14 m

∴ Volume of cylinder = πr2h = 22

7×(14)2×(19 − 14)

= 3080 m3

S31. Ans.(e)

Sol. From Statement I,

Total age of D and E = 14 yr…(i)

From Statement II,

Average age of A, B, C and F = 50 yr

∴ Total age of A, B, C and F = 4 × 50 = 200 yr…(ii)

From statement III,

Average age of A,B,D and E = 40 yr.

∴ Total age of A,B,D and E = 160 yr.

Either From Eqs. (i) and (ii), we can find the total age of A, B, C, D, E and F and then their average age

or from eq.(i), (ii) and (iii) together we can find the desired answer.

S32. Ans.(d)

Sol. Area of right angled triangle = 1

2 × Base × Height

If any two sides of a right angled triangle are known, then third side can be found out. Hence, any

two of the three statements are sufficient to find the area of the triangle.

S33. Ans.(c)

Sol. From statement I,

(A+B)’s one day’s work = 1

8

From Statement II,

(B + C)’s one day’s work = 1

10

From Statement III,

(A + B) one day’s work = 1

12


On adding all the three questions, we get

2(A + B + C)’s one day’s work

=1

8+

1

10+

1

12=

37

120

∴ (A + B + C)’s one day’s work

=37

240

Hence, we can calculate the time taken by B to complete the work.

S34. Ans.(e)

Sol. From Statement I,

SI =P×R×T

100⇒ P =

P×R×T

100

⇒ R =100

T=

100

10= 10%

From Statement II,

For 2 yr, CI − SI =PR2

1002

150 =15000×R2

10000

⇒ R2 = 100 ⇒ R = 10%

S35. Ans.(c)

Sol. Let marks scored by Gapplu in English = x

Then, from Statement III,

Marks in Science = x + 12

From Statements I and III,

Marks in Mathematics = x + 12 + 20 = x + 32

From Statements II and III,

Marks in (Mathematics + Science + English) = 198

∴ x + 32 + x + 12 + x = 198 ⇒ 3x = 198 − 44 ⇒ 3x = 154 ⇒ x = 511

3

Thus, all the Statements are required to answer the question.

S36. Ans.(a)

Sol.

Clearly replace 35 → 37


S37. Ans.(e)

Sol.

6 × 2 + 2 = 14

Replace 12 with 14

S38. Ans.(b)

Sol.

102 + 125 = 227

Replace 229 → 227

S39. Ans.(c)

Sol.

S40. Ans.(a)

Sol.

Replace 65 → 70

S41. Ans.(c)

Sol. Total tourist in year 2017 in U.P = 20×16

100×

5

4= 4 Lakhs

And we know percentage of male and female tourist is same in 2017 as in 2016 for U.P.

∴ Required difference =30

100×4,00,000 = 1,20,000

S42. Ans.(a)

Sol. Tourist visiting in year 2016 in J&K = 20×23

100= 4.6 Lakh

Who can speak both Hindi and Urdu =7

10×4.6 = 3.22 Lakh


S43. Ans.(b)

Sol. Tourists who left Haryana = 20×15

100×

20

100= 0.6 Lakh

Female in Punjab in year 2016 = 20×14

100×

20

100= 0.56 Lakh

Male in Punjab in year 2016 = 20×14

100×

80

100 = 2.24 Lakh

After increase male in Punjab = 2.24 + 0.6 = 2.84 Lakh

∴ Required ratio =2.84

0.56=

284

56= 71 ∶ 14

S44. Ans.(c)

Sol. Since percentage of male and female tourist of year 2017 is not given.

S45. Ans.(d)

Sol. Required percentage

=20×

11

100×

85

100

20×21

100×

25

100

×100 = 178.095 ≈ 178%

Solutions (46-50):

Let the quantity of Rasgulla, Rasmalai and Kalakand be 6x, 10x and 9x respectively.

Total quantity of Kalakand =18900

420= 45 kg

∴ Total quantity of Rasgula = 45×6

9= 30 kg

Total quantity of Rasmalai = 45×10

9= 50 kg

Now, S.P. of Kalakand

=100 +

275

21

100×420 = Rs. 475/kg

∴ M. P. of Kalakand = 475×100

95

= Rs. 500 kg⁄

S. P. of Rasmalia =90

100×500

= Rs. 450 kg⁄

C. P. of Rasgulla =[46400 – (50×400) – (45×420)]

30

= Rs. 250 kg⁄

Profit per kg of Rasgulla

=5875 – (50×50) – (45×55)

30

= Rs. 30


∴ S.P. per kg of Rasgulla = 250 + 30 = Rs. 280

And M.P. per kg of Rasgulla

=140

100×250 = Rs. 350

Sweets Quantity

(Kg)

C.P.

(in Rs/kg)

M.P.

(Rs./kg)

S.P.

(Rs./kg)

Profit

(Rs./kg)

Rasgulla 30 250 350 280 30

Rasmalai 50 400 500 450 50

Kalakand 45 420 500 475 55

S46. Ans.(b)

Sol. Required average C.P. per kg =46400

125= Rs. 371.2

S47. Ans.(d)

Sol. New S.P. = 80

100× 475 = Rs. 380/kg

∴ Loss% = 40

420 × 100 = 9

11

21 %

S48. Ans.(c)

Sol. Total sweets bought = 30 + 50 + 45 = 125 kg

S49. Ans.(a)

Sol. Total. C.P. = 50 × 400 = Rs. 20,000

Total S.P. = 40 × 450 = Rs. 18,000

∴ Required loss% =2000

20000×100 = 10%

S50. Ans.(b)

Sol. Required percentage =80

500×100

= 16%

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