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1 Adda247 | No. 1 APP for Banking & SSC Preparation Website: bankersadda.com | sscadda.com | store.adda247.com | Email: [email protected]
2 Adda247 | No. 1 APP for Banking & SSC Preparation Website: bankersadda.com | sscadda.com | store.adda247.com | Email: [email protected]
Solutions S1. Ans.(e) Sol. 1
A+
1
B=
1
72;
1
B+
1
C=
1
120
and 1
C+
1
A=
1
90
∴ 2 (1
𝐴+
1
𝐵+
1
𝐶) = (
1
72+
1
120+
1
90) ⇒ (
1
𝐴+
1
𝐵+
1
𝐶) =
1
2[
5 + 3 + 4
6 × 5 × 12] =
1
60
i.e. A, B and C together can do the work in 60 days. S2. Ans.(c) Sol.
Remaining work = 153 Bharat + Priyanka = 17 unit/day
∴ Required time =153
17= 9 days
S3. Ans.(a) Sol. Due to reduced efficiency, their per day work is
90
100 × 9=
1
10 for A and
72
100 × 18=
1
25 for B
∴ Time taken to complete the work =1
1
10+
1
25
=50
7 i, e, 7
1
7days
S4. Ans.(b)
Sol. Time taken by P to cultivate 4
5th of the land = 12 days
Time taken by Q to cultivate 4
5th of the land = 10 × 3 ×
4
5= 24 days
Time taken by P and Q together to cultivate 4
5th of land = (
11
12+
1
24
)
= 8 days S5. Ans.(e)
Sol. Work done in 8 days = 8 ×1
12=
2
3
Remaining work =1
3
Per day work of 25 men =1
3×
1
6=
1
18
Per day work of 15 women =1
12−
1
18=
1
36
Number of days taken by 15 women to complete the work = 36 days
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S6. Ans.(a)
Sol. Let the speed of X be a kmph. Distance travelled by X in 2 hours = 2a km.
Suppose X takes t hours to travel 1𝑡ℎ
5 of the distance AB.
Y would take (t–2) hours to travel 1𝑡ℎ
5 of the distance AB.
As Y’s speed is thrice that of X’s speed. 𝑡−2
𝑡=
1
3
t = 3 1𝑡ℎ
5 of the distance AB = 3a km.
AB =15a km
Time taken by x to cover 30a km = 15𝑎
𝑎= 15 hours
Time taken by Y to cover 30a km = 15𝑎
3𝑎= 5 ℎ𝑜𝑢𝑟𝑠.
∴ Difference in the times = 10 hours.
S7. Ans.(d)
Sol. Speed of train =Length of train and platform
time taken in crossing
We do not know the length of platform.
S8. Ans.(b)
Sol. Let he covers x km by bus.
∴ 157
)–140(
14=+
xx
x + 280 – 2x = 210
x = 70 km
Distance covered by cycle = 140 – 70 = 70 km
S9. Ans.(c)
Sol. Volume of sphere = 4
3𝜋𝑟3
Where, r = radius of sphere
∴ 4
3𝜋𝑟3 =
4000𝜋
3
⇒ r³ = 1000
⇒ r = 10 m
∴ Radius of circle = 3
5× 10 = 6 m
ATQ, 𝑎²
𝜋𝑟²=
14
11 ; a = side of square
⇒ 7
22 ×
𝑎2
36 =
14
11
⇒ a² = 4 × 36
⇒ a = 12 m
∴ perimeter = 48 m
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S10. Ans.(c)
Sol. Radius of circular field = side of square field = a (let)
ATQ,
2πa – 4a = 32
⇒ a =32×7
(44−28) = 14 m
∴ Perimeter of square field = 56 m
Solutions (11-15):
Given, Commission received by the distributor = 7000 Rs.
So, the number of bottles sold by distributor = 7000
1000 × 50 = 350
Total number of bottles received by him in the whole stock to sell = 350 + 40 = 390
Production cost of each bottle = 780000
390 = 2000 Rs.
Marked price of each bottle = 2000 × 1.3 = 2600 𝑅𝑠.
Total selling price of 350 bottles = 350 × 2000 + 140000 = 840000 Rs.
Selling price of each bottles = 840000
350 = 2400 Rs.
Discount allow by distributor (y) = 2600−2400
2600 × 100 =
200
2600 × 100 = 7
9
13%
S11. Ans.(b)
Sol. Y = 7 9
13%
S12. Ans.(d)
Sol. Required ratio = 100
13
390 =
100
13 ×390 = 10: 507
S13. Ans.(a)
Sol. New selling price of one bottle = 2600 ×90
100= 2340
Required profit % = 2340−2000
2000 × 100 = 17%
S14. Ans.(b)
Sol. Total stock which another distributor sold = (350 + 450) = 800
Total commission received by another distributor
= 800
50 × 1000 = 16000 Rs.
New cost price of one bottle = 2000 + 16000
800 = 2020 Rs.
S15. Ans.(a)
Sol. Selling price = 2600 ×95
100 ×
7
8= 2161.25 Rs.
Required profit = 2161.25 −2000 = 161.25 𝑅𝑠.
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S16. Ans.(c)
Sol. 75% × 450 + 25% × 850 = ?
? =25
100[3 × 450 + 850] =
1
4[2200] = 550
S17. Ans.(a)
Sol. ? = √273 − 357 + 280 = √196 = 14
S18. Ans.(e)
Sol.
(4)? =32 × 512 × 8
128= 1024
(4)? = (4)5
⇒ ? = 5
S19. Ans.(c)
Sol.
? + 2 +1
3+ 5 +
1
6 – 7 –
8
11= 3 +
3
11+ 8 +
1
2
? = 11 +3
11+
8
11+
1
2 –
1
3 –
1
6
= 12 +3 – 2 – 1
6= 12
S20. Ans.(e)
Sol.
81
3% × 240 + 8
1
3% × 384 =
25
100×?
?
4=
25
300[624]
? = 208
Solutions (21-24):
Total executives = 1200
Subscribed for Time magazine = 880
Subscribed for Economy = 650
No. of executives subscribed for both = (880 + 650) – 1200 = 330
S21. Ans.(a)
Sol. Required probability =880
1200=
11
15
S22. Ans.(c)
Sol. Required probability = 650
1200=
13
24
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S23. Ans.(b)
Sol. Required probability =330
1200=
11
40
S24. Ans.(a)
Sol. Required probability =330
880=
3
8
S25. Ans.(b)
Sol. Required probability
=1
3×
2
5×
1
5+
2
5×
4
5×
2
3+
1
3×
3
5×
4
5+
1
3×
2
5×
4
5
=2+16+12+8
75 =
38
75
S26. 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
S27. 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
S28. Ans.(b)
Sol.
I. 8x² + 18x + 9 = 0
⇒ 8x² + 12x + 6x + 9 = 0
⇒ (2x + 3) (4x + 3) = 0
⇒ x = –3/2, –3/4
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II. 4y² + 19y + 21 = 0
⇒ 4y² + 12y + 7y + 21 = 0
⇒ (y + 3) (4y + 7) = 0
⇒ x = –3, –7/4
x >y
S29. 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
S30. Ans.(b)
Sol.
I. 16x² + 20x + 6 = 0
⇒ 8x² + 10x + 3 = 0
⇒ 8x² + 4x + 6x + 3 = 0
⇒ (2x + 1) (4x + 3) = 0
⇒ x = –1/2, –3/4
II. 10y² + 38y + 24 = 0
⇒ 5y² + 19y + 12 = 0
⇒ 5y² + 15y + 4y + 12 = 0
⇒ (y + 3) (5y + 4) = 0
y = –3, –4/5
x > y
S31. Ans.(a)
Sol. Quantity 1: Let the maximum marks be x.
∴ (65-8) % of x = 684
⇒ 𝑥 ×57
100= 684
⇒ 𝑥 =684 × 100
57= 1200
Quantity 2: The word VIRTUAL consists of 7 distinct letters in which
vowels are A, I, U
∴ Required number of arrangements = 5! × 3!
= 5 × 4 × 3 × 2 × 1 × 3 × 2 × 1 = 720
quantity I > quantity II
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S32. Ans.(b) Sol. Quantity 1: Let Number of spherical balls made = 𝑛 Volume of Cylinder = Volume of spherical ball × 𝑛
𝜋𝑟2ℎ =4
3𝜋𝑟3 × 𝑛
𝜋 × 62 × 24 =4𝜋
3× 33 × 𝑛
𝑛 = 24
Quantity 2: (A + B + C)’s one day work = 1
2(
1
12+
1
15+
1
20) =
1
10
A’s one day work = 1
10−
1
15=
1
30
∴ A can complete the work in 30 days alone. Quantity 1 < Quantity 2 S33. Ans.(a)
Sol. Quantity 2: Cost price for retailer = 32.76 ×5
7= 23.4
Cost price for manufacturer = 23.4 ×100
125×
100
117 = 16
Quantity 1: Let area of hall = x 𝑚2 ∴ total material cost = 250 x Labor cost = Rs. 3500 ∴ 250x + 3500 = Rs. 14500
X = 11000
250= 44 𝑚2
Quantity 1 > Quantity 2 S34. Ans.(b) Sol. Case I
4800 = 3600 [1 +R
100]
3
(1 +R
100)
3
=4
3 … . (i)
Now, Population after 3 years will be
4800 [1 +R
100] 3 = 4800 ×
4
3= 6400
S35. Ans.(a) Sol. 30
U+
44
D= 10 … . (i)
40
U+
55
D= 13 … . (ii)
Apply (i) × 40 – (ii) × 30
⇒1760
D−
1650
D= 400 − 390
⇒110
D= 10 ⇒ D = 11 kmph
∴ U = 5 kmph
Rate of current =1
2(D − U) = 3 kmph
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S36. Ans.(b)
Sol. Required answers
= 50
300 × 60,000 +
25
100 × 80,000 +
20
100 × 70,000
= 10,000 + 20,000 + 14,000 = 44,000
S37. Ans.(c)
Sol. Total no. of boys from collage Q = 2
5 × 75,000 = 30,000
Total no. of girls from college S = 4
9 × 90,000 = 40,000
∴ Required percentage = 40,000−30,000
40,000 × 100 = 25% Less
S38. Ans.(a)
Sol. Required Average
= 1
3× (
50
300× 60,000 +
40
100× 75,000 +
25
100× 80,000
= 1
3× (10,000 + 30,000 + 20,000)
= 1
3 × 60,000
= 20,000
S39. Ans.(d)
Sol. Total no. of boys from collages R and S together
= 5
8 × 80,000 +
5
9 × 90,000
= 50,000 + 50,000 = 1,00,000
Total no. of girls from colleges R & S together
= 3
8 × 80,000 +
4
9× 90,000
= 30,000 + 40,000 = 70,000
∴ Required difference = 1,00,000 – 70,000 = 30,000
S40. Ans.(a)
Sol. Required no. of sport students who are not sent for state level games
= 80
100×
50
300× 60,000 +
75
100×
20
100× 70,000
= 8,000 + 10,500
= 18,500
S41. Ans.(a)
Sol. Required average no.
=1
5× (5 + 10 + 15 + 25 + 35) × 1000
=1
5× 90 × 1000
= 18000
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S42. Ans.(e)
Sol. Students in college A (10+ 15+ 15 + 20 + 30) = 90 thousand
Students in college B = (5 + 10 + 15 + 25 + 35) = 90 thousand
So, least no. of students is in colleges A and B both
S43. Ans.(b)
Sol. Required percentage
=25 − 20
20× 100
= 25% 𝑚𝑜𝑟𝑒
S44. Ans.(c)
Sol. Required difference
= (15 + 20 + 25 + 35 + 30) – (5 + 10 + 15 + 35 + 25)
= 125 – 90
= 35 thousand
S45. Ans.(d)
Sol. Required ratio
=(10 + 15 + 15 + 20 + 30)
(5 + 10 + 15 + 35 + 25)
= 1 : 1
S46. Ans.(b)
Sol. Central angle for Class VI
= 360 – (106.8 + 91.2 + 71.2 + 43.2 + 26.4) = 21.2°
∵ 21.2° → 848
∴ 360° → 848 ×360
21.2= 14400
∴ Total chocolates = 14400
∴ Required percentage = 26.4
360×14400
848× 100 ≃ 125%
S47. Ans.(c)
Sol. Total Chocolates distributed in class I = 106.8
21.2× 848 = 4,272
∴Required no. of boys =3
4×
4272
2= 1602
S48. Ans.(d)
Sol. Required average = 1
2×
(91.2 + 71.2 )
21.2× 848 =
6496
2= 3248
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S49. Ans.(b)
Sol. Total chocolates distributed among students of class VI and I
together = 106.8
21.2× 848 + 848 = 5120
And, that among class II and IV
= 43.2
21.2 × 848 +
91.2
21.2× 848 = 5376
∴ Required percentage = 5376 – 5120
5120× 100 = 5% 𝑚𝑜𝑟𝑒
Or
chocolates distributed among students in class VI
360° - (91.2° + 71.2° + 43.2° +26.4° + 106.8°) = 21.2°
Required percentage
=(91.2° + 43.2°) − (106.8° + 21.2°)
(106.8° + 21.2°)× 100
=134.4°−128°
128°× 100 = 5%
S50. Ans.(e)
Sol. Required difference = (106.8 + 91.2 + 21.2)
21.2× 848 ~
(71.2 + 43.2 + 26.4)
21.2× 848
= 8768 ~ 5632 = 3136
