# Quantitative Aptitude Quiz For IBPS SO Pre: 7th December 2018

**Dear Aspirants,**

Numerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.

**Directions (1-5): In the following Questions, two equations I and II are given. You have to solve both the equations and Give answer**

**Q1. (i) x² – 16x + 63 = 0**

**(ii) y² – 22y + 117 = 0**

If x > y

if x < y

if x ≤ y

if x ≥ y

if x = y or the relationship between x and y cannot be established.

Solution:

(i) x² – 16x + 63 = 0

x² – 7x – 9x + 63 = 0

(x – 7) (x – 9) = 0

x = 7, 9

(ii) y² – 22y + 117 = 0

y² – 9y – 13y + 117 = 0

(y – 9) (y – 13) = 0

y = 9, 13

y ≥ x

x² – 7x – 9x + 63 = 0

(x – 7) (x – 9) = 0

x = 7, 9

(ii) y² – 22y + 117 = 0

y² – 9y – 13y + 117 = 0

(y – 9) (y – 13) = 0

y = 9, 13

y ≥ x

**Q2. (i) x² – 3x – 18 = 0**

**(ii) 2y² + 11y + 14 = 0**

If x > y

if x < y

if x ≤ y

if x ≥ y

if x = y or the relationship between x and y cannot be established.

Solution:

(i) x² – 3x – 18 = 0

x² - 6x + 3x – 18 = 0

x (x – 6) + 3 (x – 6) = 0

(x – 6) (x + 3)

x = –3, 6

(ii) 2y² + 11y + 14 = 0

2y² + 7y + 4y + 14 = 0

y (2y + 7) + 2 (2y + 7) = 0

(y + 2) (2y + 7) = 0

y=–7/2, –2 No relation between x & y

x² - 6x + 3x – 18 = 0

x (x – 6) + 3 (x – 6) = 0

(x – 6) (x + 3)

x = –3, 6

(ii) 2y² + 11y + 14 = 0

2y² + 7y + 4y + 14 = 0

y (2y + 7) + 2 (2y + 7) = 0

(y + 2) (2y + 7) = 0

y=–7/2, –2 No relation between x & y

**Q3. (i) x² – 23x + 120 = 0**

**(ii) y² – y – 42 = 0**

If x > y

if x < y

if x ≤ y

if x ≥ y

if x = y or the relationship between x and y cannot be established.

Solution:

(i) x² – 23x + 120 = 0

x² – 15x – 8x + 120 = 0

x (x – 15) – 8 (x – 15) = 0

(x – 8) (x – 15) = 0

x = 8, 15

(ii) y² – y – 42 = 0

y² – 7y + 6y – 42 = 0

y (y – 7) + 6 (y – 7) = 0

(y + 6) (y – 7) = 0

y = –6, 7

x > y

x² – 15x – 8x + 120 = 0

x (x – 15) – 8 (x – 15) = 0

(x – 8) (x – 15) = 0

x = 8, 15

(ii) y² – y – 42 = 0

y² – 7y + 6y – 42 = 0

y (y – 7) + 6 (y – 7) = 0

(y + 6) (y – 7) = 0

y = –6, 7

x > y

**Q4. (i) 2x² – 13x + 15 = 0**

**(ii) 2y² + 11 y – 21 = 0**

If x > y

if x < y

if x ≤ y

if x ≥ y

if x = y or the relationship between x and y cannot be established.

Solution:

(i) 2x² – 13x + 15 = 0

2x² – 10x – 3x + 15 = 0

2x (x – 5) – 3 (x – 5) = 0

(2x – 3) (x – 5) = 0

x = 3/2, 5

(ii) 2y² + 11 y – 21 = 0

2y² + 14y – 3y – 21 = 0

2y (y + 7) – 3 (y + 7) = 0

(2y – 3) (y + 7) = 0

y=–7, 3/2

x ≥ y

2x² – 10x – 3x + 15 = 0

2x (x – 5) – 3 (x – 5) = 0

(2x – 3) (x – 5) = 0

x = 3/2, 5

(ii) 2y² + 11 y – 21 = 0

2y² + 14y – 3y – 21 = 0

2y (y + 7) – 3 (y + 7) = 0

(2y – 3) (y + 7) = 0

y=–7, 3/2

x ≥ y

**Q5. (i) x² = 36**

**(ii) y³ = 216**

If x > y

if x < y

if x ≤ y

if x ≥ y

if x = y or the relationship between x and y cannot be established.

Solution:

(i) x² = 36

x = –6, +6

(ii) y³ = 216

y = 6

x ≤ y

x = –6, +6

(ii) y³ = 216

y = 6

x ≤ y

**Directions (6-10): What should come in place of question mark (?) in the following given questions (Note: You need not to calculate the exact value.)?**

**Q6. 2.99/6.98÷291.01/238.02×387.98+46.02% 444.98= ?**

320

361

381

241

341

Solution:

?=3/7×238/291×388+46/100×445

=136 + 204.7

= 341

=136 + 204.7

= 341

**Q7. (2.87)²+(11.87)²+(9.08)²+(6.89)²=?**

283

251

315

213

173

Solution:

9+144+81+49=283

**Q8. √288.98×16.99+1083.11÷56.89=?**

364

308

393

345

413

Solution:

=17×17+19

=289+19

=308

=289+19

=308

**Q9. 94.89×12.93+√36.08×14.94=53.06×√(?)**

144

625

441

729

961

Solution:

1235+6×15=53×√x

√x=25

x=625

√x=25

x=625

**Q10. (239.98 % of 639.93)÷(7.94)² =?**

36

45

39

24

69

Solution:

240/100×640×1/64=24

**Directions (11-15): Study the bar graph to answer the following questions. The given bar graph shows the total investment (in Rs. thousand) and Percentage of Richa's investment out of total Richa’s and Diksha’s investment in 6 schemes (A, B, C, D, E and F)**

**Q11. Scheme A offers simple interest at a certain rate of interest (per cent per annum). If the difference between the interest earned by Richa and Diksha from scheme A after 4 yr is Rs. 4435.20, then what is the rate of interest (per cent per annum)?**

17.5

18

16.5

20

15

Solution:

Amount invested by Richa in scheme A = 54% of 84000 = Rs. 45360

∴ Amount invested by Diksha in scheme A = 84000 – 45360 = Rs. 38640

Let the required rate be r% per annum. Then,

= (45360×r×4)/100-(38640×r×4)/100 = 4435.20

⇒6720×r×4 = 443520

⇒ r = 16.5%

∴ Amount invested by Diksha in scheme A = 84000 – 45360 = Rs. 38640

Let the required rate be r% per annum. Then,

= (45360×r×4)/100-(38640×r×4)/100 = 4435.20

⇒6720×r×4 = 443520

⇒ r = 16.5%

**Q12. What is the respective ratio between total amount invested by Richa in schemes C and E together and total amount invested by Diksha in the same scheme together?**

31 : 44

31 : 42

27 : 44

35 : 48

29 : 38

Solution:

Required ratio = (Total amount invested by Richa in schemes C and E together): (Total amount invested by Diksha in schemes C and E together)

= (40% of 32000 + 42% of 64000): (60% of 32000 + 58% of 64000)

= 39680: 56320 = 31: 44

= (40% of 32000 + 42% of 64000): (60% of 32000 + 58% of 64000)

= 39680: 56320 = 31: 44

**Q13. If scheme C offers compound interest (compounded annually) at 12% per annum, then what is the difference between interest earned by Richa and Diksha from scheme C after 2 years?**

Rs. 1628.16

Rs. 1584.38

Rs. 1672.74

Rs. 1536.58

Rs. 1722.96

Solution:

Difference of amount invested by Richa and Diksha in

Scheme C = 60% of 32000 – 40% of 32000 = 20% of 32000 = Rs. 6400

∴ Required difference in their interest

=6400[(1+12/100)²-1]=6400×0.2544=Rs.1628.16

Scheme C = 60% of 32000 – 40% of 32000 = 20% of 32000 = Rs. 6400

∴ Required difference in their interest

=6400[(1+12/100)²-1]=6400×0.2544=Rs.1628.16

**Q14. Diksha invested in scheme F for 4 yr. If scheme F offers simple interest at 7% per annum for the first two years and then compound interest at 10% per annum (compound annually) for the 3rd and 4th year, then what will be the interest earned by Diksha after 4 yr?**

Rs. 13548.64

Rs. 13112.064

Rs. 12242.5

Rs. 12364

Rs. 11886

Solution:

Amount invested by Diksha in investment F

= (100 – 64) % of 96000 = 36% of 96000 = Rs. 34560

Then, total interest earned by Diksha after 4 years

= (34560×7×2)/100+(34560+SI of first 2 years)[(1+10/100)²-1]

= 4838.40 + 8273.664 = Rs. 13112.064

= (100 – 64) % of 96000 = 36% of 96000 = Rs. 34560

Then, total interest earned by Diksha after 4 years

= (34560×7×2)/100+(34560+SI of first 2 years)[(1+10/100)²-1]

= 4838.40 + 8273.664 = Rs. 13112.064

**Q15. Amount invested by Richa in scheme G is equal to the amount invested by her in scheme B. The rate of interest per annum of schemes G and B is same. The only difference is scheme G offers compound interest (compounded annually), whereas the scheme B offers simple interest. If the difference between the interest earned by Richa from both the schemes after 2 yr is Rs. 349.92, then what is the rate of interest?**

9%

5%

13%

11%

7%

Solution:

Amount invested by Richa in each of scheme G and B

= 60% of 72000 = 43200

Let the rate of interest be r% per annum.

Then, according to the question,

C.I -S.I. = PR²/100² (for two years)

349.92= 43200×r²/100²

or, r²=81

∴ r = 9%

= 60% of 72000 = 43200

Let the rate of interest be r% per annum.

Then, according to the question,

C.I -S.I. = PR²/100² (for two years)

349.92= 43200×r²/100²

or, r²=81

∴ r = 9%

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