SBI Clerk 2022 Day 3 Numerical Ability Online Test 3

SBI Clerk 2022  Day 3 Numerical Ability Online Test 3

 

1.  Directions(Q.1 to Q.5 ):

Identify which number is wrong in the given series.

3, 4, 10, 32, 136, 685, 41

a

   32

b

   4116

c

   10

d

   136

e

   None of these

View Answer

Answer & Explanation

Answer : Option A
Explanation :

2. 

2, 3, 4, 4, 6, 8, 9, 12, 16.

a

   9

b

   6

c

   12

d

   3

e

   None of these

View Answer

Answer & Explanation

Answer : Option A
Explanation :

3. 

11, 5, 20, 12, 40, 26, 74, 54

a

   26

b

   20

c

   40

d

   5

e

   None of these

View Answer

Answer & Explanation

Answer : Option C
Explanation :
There are two series in the given series :

Hence the wrong term is 40.

4. 

24576, 6144, 1536, 386, 96, 4

a

   386

b

   1536

c

   96

d

   6144

e

   None of these

View Answer

Answer & Explanation

Answer : Option A
Explanation :

The succeeding numbers are obtained by dividing the preceding numbers by 4. Therefore, the number 386 does not fit in the series and must be replaced by 384.

5. 

69, 55, 26, 13, 5

a

   5

b

   26

c

   55

d

   13

e

   None of these

View Answer

Answer & Explanation

Answer : Option A
Explanation :

Thus, 5 does not fit in the series and should be replaced by 4.

6. 

Mrs. X spends `535 in purchasing some shirts and ties for her husband. If shirts cost` 43 each and the ties cost ` 21 each, then what is the ratio of the shirts to the ties, that are purchased ?

a

   3 : 4

b

   2 : 1

c

   2 : 3

d

   1 : 2

e

   None of these

View Answer

Answer & Explanation

Answer : Option B
Explanation :

Mrs. X spends = ₹ 535

$\therefore$ Total cost $=43$ shirt $+21$ ties $=535$

By hit and trial, $\mathrm{S}=10, \mathrm{~T}=5$

$\Rightarrow$ Total cost $=43 \times 10+21 \times 5=535$

Hence, Ratio of shirts to ties $=10: 5=2: 1$

7. 

A reduction of $20 \%$ in the price of sugar enables a purchaser to obtain $2 \frac{1}{2} \mathrm{~kg}$ more for $₹ 160$. Find the original price per $\mathrm{kg}$ of sugar.

a

   ` 12

b

   ` 20

c

   ` 18

d

   ` 16

e

   None of these

View Answer

Answer & Explanation

Answer : Option D
Explanation :

Total amount used for purchasing $=₹ 160$. A reduction of $20 \%$ in the price means, now a person gets $5 / 2 \mathrm{~kg}$ for $₹ 32$ and this is the present price of the sugar.

$\therefore$ Present price per $\mathrm{kg}=\frac{32}{5} \times 2=₹ 12.8$

Let the original price be $₹ \mathrm{x}$. Then new price is arrived after reduction of $20 \%$ on it.

$\Rightarrow \mathrm{x} \times 0.8=12.8 \text { or } \mathrm{x}=₹ 16 \text {. }$

8. 

Anish spends 25% of his salary on house rent, 5% on food, 15% on travel, 10% on clothes and the remaining amount of ` 22,500 is saved. What is Anish’s salary ?

a

   `40,000

b

   `40,500

c

   `50,000

d

   `45,500

e

   None of these

View Answer

Answer & Explanation

Answer : Option C
Explanation :

Total expense percentage = (25 + 5 + 15 + 10)% = 55%

Savings % = 100 – 55 = 45%

$\because 45 \equiv 22500$

$\therefore 100 \% \equiv \frac{22500}{45} \times 100=₹ 50000$

9. 

$\frac{2}{5}$ th of Anil’s salary is equal to Bhuvan’s salary and sevenninth of Bhuvan’s salary is equal to Chandra’s salary. If the sum of the salary of all of them is` 77,000, then, how much is Bhuvan’s salary?

a

   `45,000

b

   `15,000

c

   `28,000

d

   `18,000

e

   None of these

View Answer

Answer & Explanation

Answer : Option D
Explanation :

Let Anil’s salary be $₹ x$.

$\therefore$Bhuvan’s salary $=₹ \frac{2 x}{5}$

Chandra’s salary $=₹ \frac{2 x}{5} \times \frac{7}{9}=\frac{14 x}{45}$

$\therefore$ Anil : Bhuvan : Chandra $=x: \frac{2 x}{5}: \frac{14 x}{45}=45: 18: 14$

$\therefore$Bhuvan’s salary

$=₹\left[\frac{18}{(45+18+14)} \times 77000\right]=₹ 18000$

10. 

A tap can fill an empty tank in 12 hours and a leakage can empty the whole tank in 20 hours. If the tap and the leakage are working simultaneously, how long will it take to fill the whole tank?

a

   35 hours

b

   25 hours

c

   30 hours

d

   40 hours

e

   None of these

View Answer

Answer & Explanation

Answer : Option C
Explanation :

Part of the tank filled in an hour

$=\frac{1}{12}-\frac{1}{20}=\frac{5-3}{60}=\frac{1}{30}$

Hence, the tank will be filled in 30 hours

11.  Directions(Q.11 to Q.15 ):

Study the following chart to answer the questions given below :

Proportion of population of seven villages in 1995

In 1996, the population of villages A as well as B is increased by 10% from the year 1995. If the population of village A in 1995 was 5000 and the percentage of population below poverty line in 1996 remains same as in 1995, find approximately the population of village B below poverty line in 1996.

a

   2500

b

   4000

c

   3500

d

   45000

e

   None of these

View Answer

Answer & Explanation

Answer : Option C
Explanation :

Population of village B in $1995=5000 \times \frac{16}{13} \approx 6150$

Population of village B in $1996=6150 \times \frac{110}{100}=6750$

Population below poverty line $=52 \%$ of $6750 \approx 3500$

12. 

If in 1997 the population of village D is increased by 10% and the population of village G is reduced by 5% from 1995 and the population of village G in 1995 was 9000, what is the total population of villages D and G in 1997?

a

   19200

b

   18770

c

   19770

d

   19870

e

   None of these

View Answer

Answer & Explanation

Answer : Option C
Explanation :

Population of village $\mathrm{D}$ in $1995=9,000 \times \frac{17}{15}=10,200$

Population of village D in $1997=10,200 \times \frac{110}{100}$

$=11,220$

Population of village G in $1997=9,000 \times \frac{95}{100}=8,550$

$\therefore$ Total population of village $\mathrm{D}$ and $\mathrm{G}$ in 1997 $=11,220+8,550=19,770$

13. 

The population of village C is 2000 in 1995. What will be the ratio of population of village C below poverty line to that of the village E below poverty line in that year ?

a

   207 : 76

b

   Data inadequate

c

   152 : 207

d

   76 : 207

e

   None of these

View Answer

Answer & Explanation

Answer : Option D
Explanation :

Population of village $\mathrm{C}$ below poverty line

$=2000 \times \frac{38}{100}=760$

Population of village $\mathrm{E}$ below poverty line

$=\frac{2000}{8} \times 18 \times\left(\frac{46}{100}\right)=2070$

$\therefore$ Required ratio $=\frac{760}{2070}=76: 207$

14. 

If the population of village C below poverty line in 1995 was 1520, what was the population of village F in 1995 ?

a

   4800

b

   6500

c

   6000

d

   4000

e

   None of these

View Answer

Answer & Explanation

Answer : Option B
Explanation :

Population of village F in 1995

$=1520 \times \frac{100}{38} \times \frac{13}{8}=6500$

15. 

If in 1995 the total population of the seven villages together was 55,000 approximately, what will be population of village F in that year below poverty line ?

a

   3000

b

   2500

c

   4000

d

   3500

e

   None of these

View Answer

Answer & Explanation

Answer : Option D
Explanation :

Population of village $F$ below poverty line

$=55000 \times \frac{13}{100} \times \frac{49}{100} \approx 3500$

16. 

Ashu’s mother was three times as old as Ashu, 5 years ago. After 5 years, she will be twice as old as Ashu. How old is Ashu at present?

a

   20

b

   5

c

   10

d

   15

e

   None of these

View Answer

Answer & Explanation

Answer : Option D
Explanation :

Let the present ages of Ashu’s mother and that of

Ashu be x and y, respectively.

Then, (x–5) = 3(y –5) or x – 5 = 3y – 15

or x – 3y = – 10 ...(i)

and (x + 5) = 2 (y + 5)

And x + 5 = 2y + 10 or x – 2y = 5 ...(ii)

From (i) and (ii), we have x = 35 and y = 15

Hence, the present age of Ashu = 15 years

17. 

If 15 women or 10 men can complete a project in 55 days, in how many days will 5 women and 4 men working together complete the same project ?

a

   75

b

   8

c

   9

d

   85

e

   None of these

View Answer

Answer & Explanation

Answer : Option A
Explanation :

$15 \mathrm{~W}=10 \mathrm{M}$

Now, $5 \mathrm{~W}+4 \mathrm{M}=5 \mathrm{~W}+\frac{4 \times 15}{10} \mathrm{~W}=5 \mathrm{~W}+6 \mathrm{~W}$$=11 \mathrm{~W}$

Now, 15 women can complete the project in 55 days, then 11 women can complete the same project in

$\frac{55 \times 15}{11}=75 \text { days }$

18. 

A sum was put at simple interest at a certain rate for 2 years. Had it been put at 3% higher rate, it would have fetched ` 300 more. Find the sum.

a

   `5000

b

   `6000

c

   `4600

d

   `8230

e

   None of these

View Answer

Answer & Explanation

Answer : Option A
Explanation :

Let the sum $=$Rs. $x$ and original rate $=y \%$ per annum then, New rate $=(y+3) \%$ per annum

$\therefore \frac{x \times(y+3) \times 2}{100}-\frac{x \times y \times 2}{100}=300$$x y+3 x-x y=15000$

∴ x = 5000

Thus, the sum $=₹ 5000$

19. 

A conical flask has base radius ’a’ cm and height ’h’ cm. It is completely filled with milk. The milk is poured into a cylindrical thermos flask whose base radius is ’p’ cm. What will be the height of the solution level in the flask ?

a

   $\frac{\mathrm{p}^{2}}{3 \mathrm{~h}^{2}} \mathrm{~cm}$

b

   $\frac{3 \mathrm{a}^{2}}{\mathrm{hp}^{2}} \mathrm{~cm}$

c

   $\frac{\mathrm{a}^{2} \mathrm{~h}}{3 \mathrm{p}^{2}} \mathrm{~cm}$

d

   $\frac{3 \mathrm{hp}^{2}}{\mathrm{a}^{2}} \mathrm{~cm}$

e

   None of these

View Answer

Answer & Explanation

Answer : Option C
Explanation :

Volume of the conical flask $=$ Volume of the cylindrical flask upto the required height $(x) \mathrm{cm}$

$\frac{1}{3} \pi a^{2} h=\pi p^{2} \times x \Rightarrow \mathrm{x}=\frac{\mathrm{ha}^{2}}{3 \mathrm{p}^{2}} \mathrm{~cm}$

20. 

A train is moving at a speed of 132 km/h. If the length of the train is 110 metres, how long will it take to cross a railway platform, 165 metres long ?

a

   7.5 s

b

   10 s

c

   5s

d

   15 s

e

   None of these

View Answer

Answer & Explanation

Answer : Option A
Explanation :

Speed of the train $=132 \mathrm{~km} / \mathrm{h}=\frac{132 \times 5}{18} \mathrm{~m} / \mathrm{s}$

Distance $=(110+165)=275 \mathrm{~m}$

Time required to cross the railway platform

$=\frac{275 \times 18}{132 \times 5}=7.5 \mathrm{~s}$

21.  Directions(Q.21 to Q.25 ):

Find out the approximate value which should come in place of the question mark in the following questions. (You are not expected to find the exact value. )

$\frac{(10008.99)^{2}}{10009.001} \times \sqrt{3589} \times 0.4987=?$

a

   3000000

b

   3000

c

   300000

d

   5000

e

   9000000

View Answer

Answer & Explanation

Answer : Option C
Explanation :

$?=\frac{(10008.99)^{2}}{10009.001} \times \sqrt{3589} \times 0.4987$

$=(10009)^{2} \times \sqrt{3600}=0.50$

$=10009 \times 60 \times 0.50 \approx 300000$

22. 

399.9+ 206 × 11.009= ?

a

   4666

b

   2800

c

   2400

d

   6666

e

   2670

View Answer

Answer & Explanation

Answer : Option E
Explanation :

? = 399.9 + 206 × 11.009

= 400 + (200 + 6) × 11 = 400 + 2200 + 66 = 2670

23. 

$\frac{2}{5}+\frac{7}{8} \times \frac{17}{19} \div \frac{6}{5}=?$

a

   $\frac{3}{4}$

b

   $\frac{1}{2}$

c

   $2 \frac{1}{2}$

d

   1

e

   $\frac{9}{11}$

View Answer

Answer & Explanation

Answer : Option D
Explanation :

$?=\frac{2}{5}+\frac{7}{8} \times \frac{17}{19} \div \frac{6}{5}=\frac{2}{5}+\frac{7}{8} \times \frac{17}{19} \times \frac{5}{6}$

$=\frac{2}{5}+\frac{595}{912}=0.40+0.65 \approx 1.05 \approx 1$

24. 

$\sqrt{45689}=?$

a

   210

b

   180

c

   415

d

   150

e

   300

View Answer

Answer & Explanation

Answer : Option A
Explanation :

$?=\sqrt{45689}=213.75 \approx 210$

25. 

(299.99999)³ = ?

a

   27000000

b

   2.7 × 10⁹

c

   180000

d

   9000000000

e

   2700000

View Answer

Answer & Explanation

Answer : Option A
Explanation :

? = (299.99999)³ ≈ (300)³ = 27000000

26.  Directions(Q.26 to Q.35 ):

What will come in place of the question mark (?) in the following equations ?

$\frac{?}{\sqrt{2.25}}=550$

a

   3666.66

b

   825

c

   82.5

d

   2

e

   None of these

View Answer

Answer & Explanation

Answer : Option B
Explanation :

Let $\frac{x}{\sqrt{2.25}}=550 .$ Then, $\frac{x}{1.5}=550$

$\therefore x=(550 \times 1.5)=\left(\frac{550 \times 15}{10}\right)=825$

27. 

$\frac{(3.537-0.948)^{2}+(3.537+0.948)^{2}}{(3.537)^{2}+(0.948)^{2}}=?$

a

   4.485

b

   2.589

c

   4

d

   2

e

   None of these

View Answer

Answer & Explanation

Answer : Option D
Explanation :

Given Expression $=\frac{(a-b)^{2}+(a+b)^{2}}{\left(a^{2}+b^{2}\right)}=\frac{2\left(a^{2}+b^{2}\right)}{\left(a^{2}+b^{2}\right)}=2$

28. 

If $\frac{a}{b}=\frac{4}{3}$, then $\frac{3 a+2 b}{3 a-2 b}=$ ?

a

   –1

b

   5

c

   3

d

   6

e

   None of these

View Answer

Answer & Explanation

Answer : Option C
Explanation :

Dividing numerator as well as denominator by $b$, we get:

$\frac{3 a+2 b}{3 a-2 b}=\frac{3 \times \frac{a}{b}+2}{3 \times \frac{a}{b}-2}=\frac{3 \times \frac{4}{3}+2}{3 \times \frac{4}{3}-2}=\frac{4+2}{4-2}=3$

29. 

$\frac{(272-32)(124+176)}{17 \times 15-15}=?$

a

   2.25

b

   240

c

   300

d

   0

e

   None of these

View Answer

Answer & Explanation

Answer : Option C
Explanation :

Given Expression $=\frac{240 \times 300}{240}=300$

30. 

$\frac{112}{\sqrt{196}} \times \frac{\sqrt{576}}{12} \times \frac{\sqrt{256}}{8}=?$

a

   16

b

   32

c

   12

d

   8

e

   None of these

View Answer

Answer & Explanation

Answer : Option B
Explanation :

Given Expression $=\left(\frac{112}{14} \times \frac{24}{12} \times \frac{16}{8}\right)=32$

31. 

$\sqrt{\frac{?}{196}}=\frac{72}{56}$

a

   14

b

   18

c

   212

d

   324

e

   None of these

View Answer

Answer & Explanation

Answer : Option D
Explanation :

Let $\sqrt{\frac{x}{196}}=\frac{72}{56}=\frac{9}{7}$.

Then,$\frac{x}{196}=\frac{9}{7} \times \frac{9}{7}=\frac{81}{49}$. So, $x=\frac{81 \times 196}{49}=324$.

32. 

$\frac{17.28 \div ?}{3.6 \times 0.2}=200$

a

   0.12

b

   12

c

   1.20

d

   120

e

   None of these

View Answer

Answer & Explanation

Answer : Option A
Explanation :

Let $\frac{17.28 \div x}{3.6 \times 0.2}=200$.

Then, $\frac{17.28}{x}=200 \times 3.6 \times 0.2$

$\therefore x=\frac{17.28}{200 \times 3.6 \times 0.2}=\frac{1728}{200 \times 36 \times 2}=0.12$

33. 

$\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=?$

a

   $\frac{1}{2}$

b

   1

c

   $4+\sqrt{15}$

d

   $4-\sqrt{15}$

e

   None of these

View Answer

Answer & Explanation

Answer : Option D
Explanation :

$\frac{(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})}=\frac{(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})} \times \frac{(\sqrt{5}-\sqrt{3})}{(\sqrt{5}-\sqrt{3})}=\frac{(\sqrt{5}-\sqrt{3})^{2}}{(5-3)}$

$=\frac{5+3-2 \sqrt{15}}{2}=\frac{2(4-\sqrt{15})}{2}=(4-\sqrt{15})$

34. 

$\frac{50}{?}=\frac{?}{12 \frac{1}{2}}$

a

   25

b

   4

c

   $\frac{4}{25}$

d

   $\frac{25}{2}$

e

   None of these

View Answer

Answer & Explanation

Answer : Option A
Explanation :

Let $\frac{50}{x}=\frac{x}{\left(\frac{25}{2}\right)}$ or $x^{2}=50 \times \frac{25}{2}=625$.

$\therefore x=\sqrt{625}=25$.

35. 

$\frac{117 \times 117 \times 117-98 \times 98 \times 98}{117 \times 117+117 \times 98+98 \times 98}=?$

a

   19

b

   29

c

   311

d

   215

e

   None of these

View Answer

Answer & Explanation

Answer : Option A
Explanation :

Given Expression $=\frac{\left(a^{3}-b^{3}\right)}{\left(a^{2}+a b+b^{2}\right)}$,

where $a=117, b=98$

$=\frac{(a-b)\left(a^{2}+a b+b^{2}\right)}{\left(a^{2}+a b+b^{2}\right)}=(a-b)=(117-98)=19 .$