# sbi clerk :: numerical ability :: test 6

## Home sbi clerk / numerical ability Questions and Answers

1.  Directions(Q.1 to Q.5 ):
Given table shows the data of population in 5 different parks. Study the data carefully and answer the questions.

If 20% of total population did not visit on a particular day in park A of which male population was 60% then what percent of total population in park B is male population who visited in park A?

a

45%

b

40.4%

c

39.2%

d

48.6%

e

None of these

 Answer : Option B Explanation : male population who did not visit park A $= \frac{{20}}{{100}} \times \frac{{60}}{{100}} \times 400 = 48$ Male population who visited in park A = 400−(150+48)=202 Required % = $\frac{{202}}{{500}} \times 100 = 40.4{\rm{\% }}$

2.  What is average of male population in park B, C and D?

a

343.33

b

313.33

c

323.33

d

333.33

e

353.33

 Answer : Option D Explanation : male population in park B, C & D = (500-200)+(700-350)+(800-450) = 1000 Required average = $\frac{{1000}}{3} = 333.33$

3.  By what percent female population in park D is more or less than the male population in park E?

a

15%

b

9.09%

c

11.11%

d

14.28%

e

12.5%

 Answer : Option E Explanation : Male population in park E = 900−500=400 Required % = $\frac{{450 - 400}}{{400}} \times 100 = 12.5{\rm{\% }}$

4.  What is ratio of male population in park A & D together to female population in park B & E together?

a

6∶7

b

1∶1

c

7∶6

d

5∶6

e

5∶7

 Answer : Option A Explanation : male population in park A & D = 400−150+800−450=600 Required ratio = 600∶(200+500)= 6∶7

5.  If 30 females from each park are above 80 years age then find the average of no. of females who are below or equal to the age of 80 years from all the parks.

a

295

b

285

c

300

d

280

e

290

 Answer : Option C Explanation : total female population = 150+200+350+450+500=1650 Female population above 80 years age = 30×5=150 Required average = $\frac{{1650 - 150}}{5} = 300$

6.  Directions(Q.6 to Q.15 ):

The ratio of ages of A and B 4 years ago was 5 : 3. The sum of present ages of A, B and C is 80 years. If present age of C is equal to sum of present ages of A and B. find the present age of A.

a

17 years

b

24 years

c

20 years

d

22 years

e

18 years

 Answer : Option B Explanation : let present age of A & B be x & y years respectively $\frac{{x - 4}}{{y - 4}} = \frac{5}{3}$ 3x−12=5y−20 3x=5y−8 ……… (i) Let present age of C be z years X+y+z=80 X+y=z X+y=40 …………. (ii) On solving (i) & (ii) x=24 ???????????????????? Present age of A = 24 ????????????????????

7.  The ratio of speed of boat in still water to speed of stream is 8 : 1. It takes 4 hours by boat to cover 54 km in downstream & 42 km in upstream. Find the downstream speed of boat.

a

25 kmph

b

24 kmph

c

21 kmph

d

27 kmph

e

23 kmph

 Answer : Option D Explanation : let speed of boat in still water & stream be 8x kmph & x kmph respectively ATQ, $\frac{{54}}{{8x + x}} + \frac{{42}}{{8x - x}} = 4$ $\frac{6}{x} + \frac{6}{x} = 4 \Rightarrow x = 3$ Downstream speed = 8x + x = 27 kmph

8.  Manoj gave 60% of his salary to his wife and invested rest amount in mutual funds. His wife spends 30% amount on grocery and 20% on rent. From remaining amount, she purchased gold worth Rs. 18000. Find salary of Manoj.

a

Rs 60000

b

Rs 54000

c

Rs 64000

d

Rs 58000

e

Rs 66000

 Answer : Option A Explanation : let salary of Manoj be Rs 100x Amount given to wife = $\frac{{60}}{{100}} \times 100x = {\rm{Rs}}.{\rm{\;}}60x$ ATQ, $60x \times \frac{{50}}{{100}} = 18000$ x=600 Salary of Manoj = 100x=???????? 60000

9.  The length & breadth of a rectangle is in ratio 4 : 7. If perimeter is 88 cm. find area of rectangle.

a

414 cm2

b

336 cm2

c

448 cm2

d

524 cm2

e

396 cm2

 Answer : Option C Explanation : let length & breadth of rectangle be 4x cm & 7x cm ATQ, 2(4x+7x)=88 x=4 Area of rectangle = 4x×7x=448 cm2

10.  The radius of a circle is 14 cm. what is area of another circle having radius 1.5 times of the actual circle?

a

1296 cm2

b

1386 cm2

c

1352 cm2

d

1485 cm2

e

1276 cm2

 Answer : Option B Explanation : radius of second circle = 1.5×14=21 cm Required area of circle = $\pi {r^2} = \frac{{22}}{7} \times 21 \times 21$ =1386 cm2

11.  Directions(Q.11 to Q.15 ):
In the following two equations questions numbered (I) and (II) are given. You have to solve both equations and Give answer

I. ${x^2} - 7x + 12 = 0$
II. ${y^2} - 8y + 12 = 0$

a

If x > y

b

If x ≥ y

c

If y > x

d

If y ≥ x

e

If x = y or no relation can be established

 Answer : Option E Explanation : I. ${x^2} - 7x + 12 = 0$ ${x^2} - 4x - 3x + 12 = 0$ $\left( {x - 4} \right)\left( {x - 3} \right) = 0$ $x = 3,4$II. ${y^2} - 8y + 12 = 0$ ${y^2} - 6y - 2y + 12 = 0$ $\left( {y - 6} \right)\left( {y - 2} \right) = 0$ $y = 2,6$ No relation can be established

12.  I. $2{x^2} + x - 28 = 0$
II. $2{y^2} - 23y + 56 = 0$

a

If x > y

b

If x ≥ y

c

If y > x

d

If y ≥ x

e

If x = y or no relation can be established

 Answer : Option D Explanation : I. $2{{\rm{x}}^2} + {\rm{x}} - 28 = 0$ $2{{\rm{x}}^2} + 8{\rm{x}} - 7{\rm{x}} - 28 = 0$ $2{\rm{x}}\left( {{\rm{x}} + 4} \right) - 7\left( {{\rm{x}} + 4} \right) = 0$ $\left( {2{\rm{x}} - 7} \right)\left( {{\rm{x}} + 4} \right) = 0$ $x = - 4,\frac{7}{2}$II. $2{{\rm{y}}^2} - 23{\rm{y}} + 56 = 0$ $2{{\rm{y}}^2} - 16{\rm{y}} - 7{\rm{y}} + 56 = 0$ $2{\rm{y}}\left( {{\rm{y}} - 8} \right) - 7\left( {{\rm{y}} - 8} \right) = 0$ $\left( {2{\rm{y}} - 7} \right)\left( {{\rm{y}} - 8} \right) = 0$ $y = \frac{7}{2},8$ ${\rm{y}} \ge {\rm{x}}$

13.  I. $2{x^2} - 7x - 60 = 0$
II. $3{y^2} + 13y + 4 = 0$

a

If x > y

b

If x ≥ y

c

If y > x

d

If y ≥ x

e

If x = y or no relation can be established

 Answer : Option E Explanation : I. $2{{\rm{x}}^2} - 7{\rm{x}} - 60 = 0$ $2{{\rm{x}}^2} - 15{\rm{x}} + 8{\rm{x}} - 60 = 0$ ${\rm{x}}\left( {2{\rm{x}} - 15} \right) + 4\left( {2{\rm{x}} - 15} \right) = 0$ $\left( {{\rm{x}} + 4} \right)\left( {2x - 15} \right) = 0$ $x = - 4,\frac{{15}}{2}$II. $3{{\rm{y}}^2} + 13{\rm{y}} + 4 = 0$ $3{{\rm{y}}^2} + 12{\rm{y}} + {\rm{y}} + 4 = 0$ $3{\rm{y}}\left( {{\rm{y}} + 4} \right) + 1\left( {{\rm{y}} + 4} \right) = 0$ $\left( {3{\rm{y}} + 1} \right)\left( {{\rm{y}} + 4} \right) = 0$ $y = - \frac{1}{3}, - 4$ no relation can be established

14.  I. ${x^2} - 17x - 84 = 0$
II. ${y^2} + 4y - 117 = 0$

a

If x > y

b

If x ≥ y

c

If y > x

d

If y ≥ x

e

If x = y or no relation can be established

 Answer : Option E Explanation : ${x^2} - 17x - 84 = 0$ ${x^2} + 4x - 21x - 84 = 0$ $\left( {x + 4} \right)\left( {x - 21} \right) = 0$ $x = - 4,21$ ${y^2} + 4y - 117 = 0$ ${y^2} - 9y + 13y - 117 = 0$ $\left( {y - 9} \right)\left( {y + 13} \right) = 0$ $y = 9, - 13$ no relation can be established

15.  I. ${x^2} = 81$
II. ${(y - 9)^2} = 0$

a

If x > y

b

If x ≥ y

c

If y > x

d

If y ≥ x

e

If x = y or no relation can be established

 Answer : Option D Explanation : ${x^2} = 81$ $x = \pm 9$ ${(y - 9)^2} = 0$ $y = 9$ Clearly, $x \le y$