1 adda247 | no. 1 app for banking & ssc preparation website · 3x² + 10x + 8 = 0 ⇒ 3x² + 6x...
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Solutions S1. Ans.(a)
Sol. Cost price of mixture = 80 ×100
150
= 160
3
Milk
Water=
160
140
=8
7
∴ Water: Milk = 7: 8
S2. Ans.(d)
Sol. Let annual rate is r%
∴ r = 100 × 294
8400 × 1 = 3.5%
∴Sum at 4%
Sum at10
3%
=
1
61
2
=1
3
∴ Sum at 4% = 1
4× 8400
= Rs. 2100
S3. Ans.(b)
Sol. Let x g sugar is added
Original sugar = 600 ×40
100
= 240 g (240+x)
(600+x)× 100 = 50
⇒ 480 + 2x = 600 + x
⇒ x = 120g
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S4. Ans.(b)
Sol. According to the law of allegation,
Land at 6% loss
Land at x% profit=
x − 10
16⇒
2
53
5
=x − 10
16
2
3=
x − 10
16⇒ x =
62
3%
S5. Ans.(c)
Sol. C.P. of mixture of sugar = 30 ×100
110=
300
11rupee kg⁄
According to law of mixture
∴sugar1
sugar2=
30
25=
6
5
∴ Quantity of sugar 1 =6
5× 30 = 36 kg
S6. Ans.(e)
Sol. Selling price of T. V. = 12000 ×88
100= 10,560
Selling price of mobile phone = 10000 ×108
100= 10800
∴ Total selling price = 10,560 + 10,800 = 21360
Total purchasing price = 12,000 + 10,000 = 22,000
∴ Loss = 22,000 − 21,360 = Rs. 640
S7. Ans.(b)
Sol. Cost price of wrist watch = 2,400 ×100
75= 3200
∴ Selling price of wrist watch when sold at 25% profit
= 3200 ×125
100= Rs. 4,000
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S8. Ans.(c)
Sol. By allegation rule,
So, ratio in which two alloy should be mix = 1 : 5
Total new alloy = 42 kg.
So, required answer = 7 kg. & 35 kg.
S9. Ans.(d)
Sol. Let total distance = d km
∴ Average speed =d
d
120+
d
120+
d
40
= 24 km hr⁄
S10. Ans.(a)
Sol. Let 5 consecutive odd numbers are x–4, x–2, x, x+2, x+4
ATQ,
(x + 3) ² – (x – 3) ² = 492
⇒ 12x = 492
⇒ x = 41
∴ smallest odd no. = 41 – 4 = 37
S11. Ans.(b)
Sol. Required no. of participants
=24
100× 450 +
20
100× 750 +
15
100× 480 +
15
100× 640 +
5
100× 360= 444
S12. Ans.(c)
Sol. No. of selected participants from Vadodara in 2006
=20
100× 640 = 128
No. of selected participants from Mumbai in 2006
=16
100× 750 = 120
∴ Required percentage =128−120
128× 100
= 6.25% less
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S13. Ans.(d)
Sol. Required average
= 1
5× (
16
100× 450 +
24
100× 750 +
20
100× 480 +
10
100× 640 +
25
100× 360)
=1
5× (72 + 180 + 96 + 64 + 90)
=502
5≈ 100
S14. Ans.(b)
Sol. Selected participants from Lucknow in 2009 & 2010 together
=(25 + 30)
100× 480 = 264
And, that from Bhopal =55
100× 360 = 198
∴ Required ratio =264
198=
4
3 ⇒ 4:3
S15. Ans.(e)
Sol. Required difference
=80
100×
20
100× 450 −
60
100×
16
100× 750 = 72 – 72 = 0
S16. Ans.(d)
Sol.
S17. Ans.(b)
Sol.
S18. Ans.(a)
Sol.
S19. Ans.(c)
Sol.
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S20. Ans.(e)
Sol.
S21. Ans.(a)
Sol. CP of one article =2100
3 = 700
Profit =20
100× 700 = Rs 140
S22. Ans.(b)
Sol. Let investment of A, B and C be x, y and z respectively.
x : y : z = 1440 : 2160 : 2640
x : y : z = 6 : 9 : 11
Required sum (A + B) =15
26× 6500 = Rs 3750
S23. Ans.(d)
Sol. Upstream speed =12
20× 60 = 36 km/hr
Downstream speed =22
22× 60 = 60 𝑘𝑚/ℎ𝑟
x -y = 36 …(i)
x + y = 60 …(ii)
from (i) & (ii)
y = 12 km/hr
S24. Ans.(d)
Sol. Let efficiency of one man and one woman be ‘m’ units/day and ‘w’ units/day respectively.
15m × 12 =18w × 20
1m = 2w
(8m + 12w) ×x = 18w × 20
(16w + 12w) ×x = 18w × 20
𝑥 =18𝑤 × 20
28𝑤 = 12
6
7 days
S25. Ans.(c)
Sol. Principal =𝑆𝐼 × 100
𝑇𝑖𝑚𝑒 × 𝑅𝑎𝑡𝑒
=240 × 100
5 × 6= Rs 800
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Solutions (26-30):
Total persons who were awarded = 103
20.6× 360 = 1800
States No. of persons who were awarded
UP 542
MP 228
Maharashtra 453
Kerala 103
West Bengal 123
Haryana 351
S26. Ans. (b) Sol. Required answer = 542 + 453 + 123 = 1118 S27. Ans. (c)
Sol. Required parentage = 454–453
453× 100 =
100
453% more
S28. Ans. (d)
Sol. Total Females of MP who were awarded = 1
3 ×228 = 76
Total females of west Bengal who were awarded = 2
3×123 = 82
∴ Required answer = 82−76
82 × 100 ≃ 7.3 % less
S29. Ans. (a)
Sol. Required average = 1
3 × (542 + 228 + 103) = 291
S30. Ans. (b) Sol. Required difference = |(542 + 103 + 123) − (228 + 453 + 351)| = 264 S31. Ans.(d) Sol.
16 ×2.4
100𝑥 = 288
𝑥 = 750 S32. Ans.(b) Sol. (−251 × 21 × (−12))
𝑥=
15813
100
𝑥 = 400 S33. Ans.(d) Sol.
[130 × 130
25× 15]
1
30= 338
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S34. Ans.(e)
Sol.
√√44944 + √52441 = √212 + 229 = √441 = 21
S35. Ans.(d)
Sol. 24.375 − 1.955 = 22.420
S36. Ans.(a)
Sol. 3 Pen + 4 Er + 5 Pa = 28
A → Pa + 0.25 = Pen + Er
B → 10 Pa + 8 Er = 42.5
C→ Pen = 1.8 Er
So, using any two of the 3 statements the cost of Eraser can be determined.
S37. Ans.(c)
Sol. A → 𝑥 + 𝑦 + 𝑥 − 𝑦 = 12, 𝑥 = 6
(𝑥 + 𝑦) − (𝑥 − 𝑦) = 6, 𝑦 = 3
𝑥2: 𝑦2 = 36 ∶ 9 or 4 ∶ 1
B → 𝑦 =𝑥
2, 𝑥2 ∶ 𝑦2 = 4 ∶ 1
C → 𝑦
𝑥=
1
2, 𝑥2: 𝑦2 = 4 ∶ 1
So, anyone of the statements is sufficient to answer the question.
S38. Ans.(e)
Sol. A → 0.82 MP = 516.6, M = 630 Rs.
B → using A, 630 = 1.25 CP, CP = 504 Rs.
C → using A
M = 630, at 10% discount, SP = 567
C.P. =567
112.5= 504 Rs.
So, using either B or C with A we can find the cost price.
S39. Ans.(d)
Sol. Volume of cone = Lateral Surface area 1
3𝜋𝑟2ℎ = 𝜋𝑟ℓ [ℓ = √ℎ2 + 𝑟2]
𝑟ℎ
3= √ℎ2 + 𝑟2
Squaring both sides 1
9=
ℎ2+𝑟2
𝑟2ℎ2
1
9=
ℎ2
𝑟2ℎ2+
𝑟2
𝑟2ℎ2
1
9=
1
𝑟2+
1
ℎ2
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S40. Ans.(a) Sol.
Area = 4 π r2 Cumulative area of the two pieces = 25% more than the by sphere. Area of 2 pieces = 1.25 × 4 π r2 = 5πr2
Extra area = πr2
Extra area = Flat surface area of two new circles that are now created circles.
Area each new circle =𝜋𝑟2
2
Let radius of new circle be r1.
Now, πr12 = 𝜋𝑟2
2
r1 = 𝑟
√2
Now, r1, h and r form a right angled triangle. h2 + r12 = r2
h2 +(𝑟
√2)
2
= r2
h = 𝑟
√2
S41. Ans.(c) Sol. I. 3x² + 10x + 8 = 0 ⇒ 3x² + 6x + 4x + 8 = 0 ⇒ (x + 2) (3x + 4) = 0 ⇒ x = –2, –4/3 II. 3y² + 7y + 4 = 0 ⇒ 3y² + 3y + 4y + 4 = 0 ⇒ (y + 1) (3y + 4) = 0 ⇒ y = –1, –4/3 y ≥ x
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S42. Ans.(e)
Sol. I. 2x² + 21x + 10 = 0
⇒ 2x² + 20 + x + 10 = 0
⇒ (x + 10) (2x + 1) = 0
⇒ x = –10, –1/2
II. 3y² + 13y + 14 = 0
⇒ 3y² + 6y + 7y + 14 = 0
⇒ (y + 2) (3y + 7) = 0
⇒ y = –2, –7/3
No relation
S43. Ans.(e)
Sol. I. 4x² – 13x + 9 = 0
⇒ 4x² – 4x – 9x + 9 = 0
⇒ (x – 1) (4x – 9) = 0
⇒ x = 1, 9/4
II. 3y² – 14y + 16 = 0
⇒ 3y² – 6y – 8y + 16 = 0
⇒ (y – 2) (3y – 8) = 0
⇒ y = 2, 8/3
No relation
S44. Ans.(b)
Sol. I. 8x² + 18x + 9 = 0
⇒ 8x² + 12x + 6x + 9 = 0
⇒ (2x + 3) (4x + 3) = 0
⇒ x = –3/2, –3/4
II. 4y² + 19y + 21 = 0
⇒ 4y² + 12y + 7y + 21 = 0
⇒ (y + 3) (4y + 7) = 0
⇒ x = –3, –7/4
x >y
S45. Ans.(a)
Sol. I. 3x² + 16x + 21 = 0
⇒ 3x² + 9x + 7x + 21 = 0
⇒ (x + 3) (3x + 7) = 0
⇒ x = –3, –7/3
II. 6y² + 17y + 12 = 0
⇒ 6y² + 9y + 8y + 12 = 0
⇒ 3y (2y + 3) + 4 (2y + 3) = 0
⇒ y = – 3/2, –4/3
y > x
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S46. Ans.(b)
Sol. Required average
=1
5× (20 + 25 + 35 + 30 + 45)
= 31 thousand
S47. Ans.(c)
Sol. Required difference
|(15 + 20 + 30 + 35 + 40) − (10 + 20 + 25 + 40 + 50)|
= 5000
S48. Ans.(a)
Sol. Required percentage
=35 − 30
35× 100
= 142
7% less
S49. Ans.(d)
Sol. Required ratio =(20 + 35)
(10 + 25)=
11
7
S50. Ans.(a)
Sol. From the graph it is clear that person B gets maximum income in 2005