Numerical Ability Test - 1 - Bank Exams MCQ

Given:
24 × 9 + 84 ÷ 14 = (?) + 6²
Concept used:
Follow the BODMAS rule according to the table given below:

Calculation:
24 × 9 + 84 ÷ 14 = (?) + 6²
⇒ 216 + 6 = (?) + 36
⇒ ? = 222 - 36 = 186
∴ The value of ? is 186.

Calculation:
Number of students learning German from school B = (625 – 330)
⇒ 295
Number of students learning French from school A = 250
Required Percentage = [(295 – 250)/250] × 100
⇒ [45/250] × 100
⇒ 18%
∴ The required percentage is 18%

Calculation:

Total number of students = (550 + 625 + 680 + 575)

⇒ 2430

Total number of students learning French = (250 + 330 + 380 + 240)

⇒ 1200

Total number of students learning German = (2430 – 1200)

⇒ 1230

Required ratio = 1230 : 1200

⇒ 41 : 40

∴ The required ratio is 41 : 40

Calculation:

Number of students learning German from school C = (680 – 380)

⇒ 300

Number of students learning German from school D = (575 – 240)

⇒ 335

Required difference = (335 – 300)

⇒ 35

∴ The required difference is 35

Calculation:

The series follows following pattern:

⇒ 20² – 1 = 399

⇒ 19² – 4 = 357

⇒ 18² – 9 = 315

⇒ 17² – 16 = 273

⇒ 16² – 25 = 231

⇒ 15² – 36 = 189

∴ The value of ? is 189

Given:

The number series:

10, 5, ?, 10, 40, 320

Concept Used:

Number series following a fixed pattern.

Calculation;

The given number series follows the pattern:

10 × 2^-1 = 5

5 × 2⁰ = 5

5 × 2¹ = 10

10 × 2² = 40

40 × 2³ = 320

∴ The number coming in place of “?” in the given number series is 5

In 90 litres of the mixture,

Milk = (3/5) × 90 = 54 litres

Water = (90 - 54) = 36 litres

Let x litres of water be added

Then, 54/(x + 36) = 2/3

⇒ 2x = 90

⇒ x = 45 litres

Given:

Percentage of females in building A = 10

No. of male in building A = 180

Calculation:

Percentage of males in building A = 90

According to question,

⇒ 90% = 180

⇒ 10% = (180/90) × 10 = 20

∴ Total no. of females in building A is 20

Mistake Points
It is just a coincidence that the sum of the percentage is 100 but if we ignore that fact then there is no issue in the pie chart. Percentage of females given in the pie chart is of respective buildings.

From I,

x² – 49 = 0

⇒ x = 7, -7

From II,

y² – 25= = 0

⇒ y = 5, -5

Comparison between x and y (via Tabulation):

∴ Relation between x and y can't be established.

Given expression is,

⇒ 90% of 270 = ‘?’% of 300

⇒

⇒ 243 = ? × 3

⇒ ? = 243/3

⇒ ? = 81

1280% of 75 ÷ 30 + ? = 80

⇒ 960 ÷ 30 + ? = 80

⇒ 32 + ? = 80

⇒ ? = 80 - 32 = 48

Follow BODMAS rule to solve this question, as per the order given below,

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated

378 ÷ 42 × ∛512 + (12)2 = ?³

⇒ 9 × 8 + 144 = ?3

⇒ 72 + 144 = ?3

⇒ 216 = ?3

∴ ? = 6

Total no. of people practicing yoga below 40 years = 128 + 309 = 437
∴ Required percentage = (437/1000) × 100 = 43.7%

No. of people below 20 = 128
No. of people above 60 = 106
∴ Required percentage = [(128 – 106)/106] × 100 = 20.75%

No. of people practising yoga below 40 = 128 + 309 = 437
No. of people practising yoga above 40 = 457 + 106 = 563
∴ Required difference = 563 – 437 = 126

Sum of ratios = 48 + 55 = 103
No. of people between 20 – 30 and 30 – 40 = No. of people between 20 – 40 = 309
∴ No. of people between 20 – 30 = (48/103) × 309 = 144

Given:

Profit (P) % = 15%

S.P. = Rs. 1,104

Discount (D) = 20%

Formula used:

S.P. = C.P. × (100 + P)/100

S.P. = M.P. × (100 - D)/100

Calculation:

S.P. = C.P. × (100 + P)/100

1104 = C.P. × 1.15

C.P. = Rs. 960

Rs. 960 is an S.P. of 2nd condition,

S.P. = M.P. × (100 - D)/100

960 = M.P. × 0.8

M.P. = 960/0.8 = Rs. 1,200

Given:

Wrong values are 83 and 78.

True value is 63 and 58.

Concept:

Total increase in marks = False value – true value

Calculation:

Let, the number of students in the class be x

As the average increases by 2

Wrong value is 83 and 78

True value is 63 and 58

The total increase in marks for x students

⇒ x × average increases by 2

⇒ x × 2 = 2x

Total increase marks by adding true value at the place of wrong value

⇒ 2x = (83 + 78) – (63 + 58)

⇒ 2x = 161 – 121

⇒ x = 20

∵ We assumed that the number of students in the class be x

∴ The number of students in the class is 20.

Alternate Method

Let A be average and x be the number of students.

Than,

Average + increments in average = (wrong value/number of students)

⇒ A + 2 = [(83 + 78)/x]

⇒ A + 2 = 161/x ----(i)

Average = (true value/number of student)

⇒ A = (63 + 58)/x

⇒ A = 121/x ----(ii)

From (i) and (ii), we get

⇒ A + 2 – A = (161/x) – (121/x)

⇒ 2 = (161 – 121)/x

⇒ 2x = 40

⇒ x = 20

∵ We assumed that the number of students in the class be x

∴ The number of students in the class is 20.

Given:

Time = 2 years

Rate of Interest = 6%

Compound Interest paid by Raj = 618

Formula:

CI = P(1+ R/100)ⁿ – P

Calculation:

CI = P(1+ R/100)ⁿ – P

⇒ P(1+ 6/100)² - P = 618

⇒ P[(1+3/50)² – 1] = 618

⇒ P[(53/50)² - 1] = 618

⇒ P[(2809/2500) – 1] = 618

⇒ P(309/2500) = 618

⇒ P = (618 × 2500/309)

⇒ P = 5000

∴ Required sum is Rs 5,000

Given:

Samya can do a work in = 15 days

Sankha can do a work in = 30 days

They work for 5 hours a day

Concept:

A can complete a work in =‘d’ days

∴ In 1 day A can do = 1/d

∴ In x day A can do = x/d

Calculation:

Samya can complete a work in = (15 × 5) = 75

Work done By Samya in 1 hr = 1/75

Sankha can complete a work in = (30 × 5) = 150

Work done By Sankha in 1 hr = 1/150

(Samya + Sankha)’s 1 hr work = (1/75) + (1/150)

⇒ (2 + 1)/150

⇒ 3/150

⇒ 1/50

Time taken by Samya and Sankha together working 10 hours daily,

⇒ 50/10 = 5 days

Time taken by the inlet pipe to fill the cistern = 6 hours

So, part of cistern filled by inlet pipe in 1 hour = 1/6

Time taken to fill the cistern with leak = 6 hours + 4 hours = 10 hours

Part of cistern filled by inlet pipe with leak in 1 hour = 1/10

Let the leak take x hours to empty the filled tank.

⇒ 4x = 60

⇒ x = 15 hours

As per the given data,

From I)

⇒ x² – 18x + 45 = 0

⇒ x² - 15x - 3x + 45 = 0

⇒ x (x -15) -3 (x - 15) = 0

⇒ (x - 3)(x - 15) = 0

∴ x = 3 and x = 15

From II)

⇒ y² – 39y + 74 = 0

⇒ y² - 37y - 2y + 74 = 0

⇒ y (y - 37) - 2 (y-37) = 0

⇒ (y - 2)(y - 37) = 0

∴ y = 2 and y = 37

When x = 3 and y = 2, then x > y

x = 3 and y = 37, then x < y

x = 15 and y = 2, then x > y

x = 15 and y = 37, then x < y

According to the BODMAS rule, the priority in which the operations should be done is:

⇒ 20% of 200 × √1225 + (207)² = (?) - 25% of 160

(?) = 20% of 200 × √1225 + (207)² + 25% of 160

(?) = 1/5 × 200 × 35 + 42849 + ¼ × 160

(?) = 40 × 35 + 42849 + 40 = 44289.

Concept used:

25% → 1/4
Calculation:
25% of 700 + 40% of 350 - 23 × 5 = ?
⇒ 1/4 × 700 + (40/100) × 350 - 23 × 5 = ?
⇒ 175 + 140 -115 = ?
⇒ 315 -115 = 200 = ?
∴ ? = 200

Given:

The sum of two numbers = 336

The HCF of two numbers = 48

Concept:

HCF is the highest common factor of two or more numbers

Calculation:

Let the two numbers are 48x and 48y respectively

According to the question,

⇒ 336 = 48x + 48y

⇒ 48 × (x + y) = 336

⇒ x + y = 7

The possible value of x and y = (2, 5) or (3, 4) because x and y are coprime numbers

When we take (x, y) = (2, 5) then,

The numbers = (48 × 2), (48 × 5) = 96, 240

∴ The required result will be "96, 240".

The logic followed here is as follows:

⇒ 3 x 4 = 12

⇒ 4 x 5 = 20

⇒ 5 x 6 = 30

⇒ 6 x 7 = 42

⇒ 7 x 8 = 56

⇒ 8 x 9 = 72

So, the wrong term is 11

Hence, the correct answer is 11.

Rules of Approximation:

1. If a number has digits to the right of the decimal less than 5, then just drop the digits to the right of the decimal. The number so obtained will be the approximated value.
2. If a number has digits to the right of the decimal more than 5, then just drop the digits to the right of the decimal and raise the remaining number by '1'.The number so obtained will be the approximated value.

Concept:
In this type of question, we are expected to calculate approximate value (not exact value), so we can replace the given numbers by their nearest perfect places which makes the calculation easy.

(143.99)^1/2 + (63.98)^1/3 = (?)^1/2

taking approximate value, we get

(144)^1/2 + (64)^1/3 = (?)^1/2

Follow BODMAS rule to solve this question, as per the order given below,

12 + 4 = √?

⇒ 16 = √?

∴ ? = 256

Given:

? ÷ 24.96% of 239.98 = 30.04

Concept:

Follow the BODMAS rule according to the table given below:

Calculation:

? ÷ (24.96%) of 239.98 = 30.04

⇒ ? ÷ 25% of 240 = 30

⇒ ? ÷ (1/4) × 240 = 30

⇒ ? ÷ 60 = 30

⇒ ? = 30 × 60

⇒ ? = 1800

∴ 1800 will come in place of the question mark (?).

Rules of Approximation:

1. If a number have digits to the right of the decimal less than 5. Then just drop the digits to the right of the decimals. The number so obtained will be the approximated value.
2. If a number have digits to the right of the decimal more than 5. Then just drop the digits to the right of the decimals and raise the remaining number by '1'.The number so obtained will be the approximated value.

Concept:
Follow BODMAS rule to solve this question, as per the order given below,

Since, we need to find out the approximate value, we can write these values to their nearest integers.
Given expression is

4968 – 159.99 × 34.99 + 616.99 =?

⇒ 4968 – 160 × 35 + 617 = ?

⇒ 4968 – 5600 + 617= ?

⇒ 4968 – 5600 + 617 = ?

⇒ 5585 – 5600 = ?

⇒ -15 = ?

∴ The value of ? is -15