SBI Clerk Mains Quantitative Aptitude (Day-44)

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Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for SBI Clerk Mains 2020 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

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Wrong number series

Directions (1 – 4): The questions below are based on the given Series – I. The series – I satisfy a certain pattern, follow the same pattern in Series – II and find the wrong number in the series –II.

1) I) 400, 200, 300, 750, 2625

II) 320, 160, 240, 610, 2100

A) 610

B) 240

C) 160

D) 320

E) 2100

2)  I) 1180, 1060, 1160, 1080, 1140, 1100

II) 2460, 2340, 2440, 2320, 2420, 2380

A) 2460

B) 2320

C) 2440

D) 2420

E) 2380

3) I) 16, 18, 39, 121, 489, 2451

II) 42, 44, 91, 275, 1113, 5571

A) 42

B) 5571

C) 91

D) 275

E) 1113

4) I) 110, 231, 131, 212, 148, 197

II) 430, 551, 451,530, 468, 517

A) 551

B) 468

C) 530

D) 451

E) 517

Application Sums

5) A, B and C together can do a work in 10 days. B & C together thrice as much as A, A & B together work 4 times as much as C. Who among the following has completed the work earlier?

i) A

ii) C

iii) B

A) Only i

B) Both i and ii

C) Only ii

D) Only iii

E) None of these

6) Initially in a mixture, the ratio of gold and copper is 5:1. ________gm of mixture is drawn and _________gm of copper is added and makes the final mixture with 70 gm gold and 26 gm copper

A) 36, 12

B) 42, 6

C) 72, 12

D) 72, 6

A) Only A

B) Only B

C) Only C

D) Only A and C

E) None of these

7) A and B’s present age ratio is 43:31. If A’s age 8 years ago and B’s age 6 years ago is in the ratio 7:5 respectively. A is _____years old 7 years ago and B is _____years old after 5 years

A) 36, 36

B) 43, 31

C) 50, 36

D) 36, 32

E) None of these

8) P, Q and R started a business by investing in the ratio of 6 : 8 : 13. After 4 months, P invested 25 % more than the initial investment and after another 5 months, P withdraws Rs. 10000 but Q invested Rs. 8000 more than the initial investment. Find the initial investment of Q, if the share of P, Q and R at the end of the year is in the ratio of 51: 68: 104?

A) Rs. 32000

B) Rs. 43000

C) Rs. 38000

D) Rs. 51000

E) None of these

9) Sangeetha invested Rs. 40000 in SI at the rate of 2x % per annum for two years and the same amount is invested in CI at double the rate of simple interest for two years, if she received Rs. 9600 more interest than SI, then find the rate of interest per annum for CI?

A) 15 %

B) 20 %

C) 10 %

D) 24 %

E) None of these

Quadratic Equation

Directions (10): Following questions contain two equations as I and II. You have to solve both equations and determine the relationship between them and give answer as,

a) If x > y

b) If x ≥ y

c) If x = y or relationship cannot be determined.

d) If x < y

e) If x ≤ y

10) I) x2 + 19x – 966 = 0

II) y2– 25x + 156 = 0

Answers :

Directions (1-4) :

1) Answer: A

Series I:

400 * 0.5 = 200

200 * 1.5 = 300

300 * 2.5 = 750

750 * 3.5 = 2625

Series II:

320 * 0.5 = 160

160 * 1.5 = 240

240 * 2.5 = 600 (Not 610)

600 * 3.5 = 2100

2) Answer: B

Series I:

1180 – 120 = 1060

1060 + 100 = 1160

1160 – 80 = 1080

1080 + 60 = 1140

1140 – 40 = 1100

Series II:

2460 – 120 = 2340

2340 + 100 = 2440

2440 – 80 = 2360 (Not 2320)

2360 + 60 = 2420

2420 – 40 = 2380

3) Answer: D

Series I:

16 * 1 + 2 = 18

18 * 2 + 3 = 39

39 * 3 + 4 = 121

121 * 4 + 5 = 489

489 * 5 + 6 = 2451

Series II:

42 * 1 + 2 = 44

44 * 2 + 3 = 91

91 * 3 + 4 = 277 (Not 275)

277 * 4 + 5 = 1113

1113 * 5 + 6 = 5571

4) Answer: C

Series I:

110 + 112 = 231

231 – 102 = 131

131 + 92 = 212

212 – 82 = 148

148 + 72 = 197

Series II:

430 + 112 = 551

551 – 102 = 451

451 + 92 = 532 (Not 532)

532 – 82 = 468

468 + 72 = 517

Directions (5-9) :

5) Answer: D

1 / a + 1 / b + 1 / c = 1 / 10- –  –  –  –  –  –  – 1

1 / b + 1 / c = 3 / a –  –  –  –  –  –  –  –  – 2

1 / a + 1 / b = 4 / c –  –  –  –  –  –  –  –  –  – 3

From 1 & 2,

1 / a + 3 / a = 1 / 10

4 / a = 1 / 10

1 / a = 1 / 40 ——- 4

From 1 & 3,

4 / c + 1 / c = 1 / 10

5 / c = 1 / 10

1 / c = 1 / 50 —— 5

Substitute 4 & 5 in 1,

1/40 + 1/b + 1/50 = 1/10

1/b = 1/10 – 1/40 – 1/50

1/b = (20 – 5 – 4)/200 = 11/200

“B” has completed the work earlier.

6) Answer: D

A) Gold in final mixture,

g – (36 * 5 / 6) =70, g = 100

Copper in final mixture,

c – (36 * 1 / 6) +12 = 26, c = 20

g:c = 5:1 A is satisfied

B)

g – (42 * 5 / 6) =70, g = 105,

c – (42 * 1 / 6) +6 = 26, c = 27

g:c = 35:9 B is not satisfied

C)

g – (72 * 5 / 6) =70, g = 130,

c – (72 * 1 / 6) +12 = 26, c = 26

g:c = 5:1 C is satisfied

D)

g – (72 * 5 / 6) =70, g = 130,

c – (72 * 1 / 6) +6 = 26, c = 32

g:c = 65:16 D is not satisfied

7) Answer: A

According to the question,

(7x + 8)/(5x + 6) = 43/31

217x + 248 = 215x + 258

2x = 10

= > x = 5

A’s age 7 years ago = (7 * 5 + 8) – 7 = 36

B’s age 5 years after = (5 * 5 + 6) + 5 = 36

8) Answer: A

The share of P, Q and R,

= > [6x * 4 + 6x * (125/100) * 5 + ((6x * (5/4)) – 10000) * 3]: [8x * 9 + (8x + 8000) * 3] : [13x * 12] = 51 : 68 : 104

= > [24x + (75x/2) + (45x/2) – 30000] : [72x + 24x + 24000] : [156x] = 51 : 68 : 104

= > [(168x – 60000) / 2]: [96x + 24000]: [156x] = 51: 68: 104

According to the question,

= > (96x + 24000) / (156x) = (68/104)

= > 96x + 24000 = 102x

= > 6x = 24000

= > x = 4000

The initial investment of Q = 8x = Rs. 32000

9) Answer: B

C.I – S.I = 9600

40000 * [(1 + (4x/100))2 – 1] – (40000 * 2x * 2)/100 = 9600

40000 * [((100 + 4x)/100)2 – 1] – 1600x = 9600

40000 * [(10000 + 800x + 16x2 – 10000)/10000] – 1600x = 9600

4 * (800x + 16x2) – 1600x = 9600

3200x + 64x2 – 1600x = 9600

64x2 + 1600x – 9600 = 0

x2 + 25x – 150 = 0

(x – 5) (x + 30) = 0

x = 5, -30 (Negative value will be eliminated)

The rate of interest per annum for CI = 4x % = 20 %

10) Answer: C

x2 – 23x + 42x – 966 = 0

x(x – 23) +42(x – 23) =0

(x – 23) (x + 42)  = 0

x= -42, 23

y2 – 13y – 12y + 156 = 0

y(y – 13) -12(y – 13) =0

(y – 12) (y – 13)  = 0

y = 12, 13

Hence, the relationship between x and y cannot be determined

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