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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%
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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
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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
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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%
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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
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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
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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
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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
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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
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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
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∴ 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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