Numerical Ability Test - 4 - Bank Exams MCQ

Concept used :

Follow the BODMAS rule to solve this question, as per the question is 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 part of the equation that contains 'Division' and 'multiplication' are calculated,

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

Given :

⇒ 64 % of 2500 + 75% of 1200 + 40% of 500 = ? × 125

⇒ 64 × 2500 / 100 + 75 × 1200 / 100 + 40 × 500 / 100 = ? × 125

⇒ 64 × 25 + 75 × 12 + 40 × 5 = ? × 125

⇒ 1600 + 900 + 200 = ? × 125

⇒ 2700 = ? × 125

⇒ ? = 2700 / 125

⇒ ? = 21.6

Hence 21.6 is the correct answer.

Given:

14 + 39 + 26 - =

Calculation:

⇒ 14 + 39 + 26 - =

⇒ (14 + 26) + 39 - =

⇒ [(14 + 26) + ( + )] + 39 - =

⇒ 40 + + 39 - 64 =

⇒ 40 + 1 + 39 - 64 =

⇒ = 16

⇒ x = 16²

⇒ x = 256

∴ The value of 'x' is 256.

Given:

I. x² – 14x + 33 = 0

II. y² – 20y + 99 = 0

Calculation:

I. x² – 14x + 33 = 0

⇒ x² – 11x – 3x + 33 = 0

⇒ x(x – 11) – 3(x – 11) = 0

⇒ (x – 11)(x – 3) = 0

Then, x = 11 or x = 3

Now,

II. y² – 20y + 99 = 0

⇒ y² – 11y – 9y + 99 = 0

⇒ y(y – 11) – 9(y – 11) = 0

⇒ (y – 11)(y – 9) = 0

Then, y = 11 or y = 9

Now,

Comparison between x and y (via Tabulation):

∴ We can clearly see that the relationship cannot be established

Let the number of girls and boys students in a school be 11a and 9a respectively.

⇒ 11a – 9a = 4000

⇒ a = 2000

Number of Girls = 11a = 22000

Number of boys = 9a = 18000

Total number of students = 22000 + 18000 = 40000

Let number of students who choose Maths and Computer be 21b and 19b respectively.

⇒ 21b + 19b = 40000

⇒ b = 1000

Number of students who choose Maths = 21b = 21000

Number of students who choose Computer = 19b = 19000

Let number of boys students who choose Maths be M.

The number of boys students who choose Computer = M – M × 50/100 = 50M/100 = M/2

Then,

⇒ 18000 = M + M/2

⇒ M = 12000

Number of boys students who choose Maths = 12000

Number of boys students who choose Computer = M/2 = 6000

Number of girls students who choose Maths = 21000 – 12000 = 9000

Number of girls students who choose Computer = 19000 – 6000 = 13000

Required difference = 13000 – 9000 = 4000

Let the number of girls and boys students in a school be 11a and 9a respectively.

⇒ 11a – 9a = 4000

⇒ a = 2000

Number of Girls = 11a = 22000

Number of boys = 9a = 18000

Total number of students = 22000 + 18000 = 40000

Let number of students who choose Maths and Computer be 21b and 19b respectively.

⇒ 21b + 19b = 40000

⇒ b = 1000

Number of students who choose Maths = 21b = 21000

Let number of boys students who choose Maths be M.

The number of boys students who choose Computer = M – M × 50/100 = 50M/100 = M/2

Then,

⇒ 18000 = M + M/2

⇒ M = 12000

Number of boys students who choose Maths = 12000

Number of girls students who choose Maths = 21000 – 12000 = 9000

Required percentage = 9000/21000 × 100

= 42.85% ≈ 43%

Let a number of girls and boys students in a school be 11a and 9a respectively.

⇒ 11a – 9a = 4000

⇒ a = 2000

Number of Girls = 11a = 22000

Number of boys = 9a = 18000

Total number of students = 22000 + 18000 = 40000

Let number of students who choose Maths and Computer be 21b and 19b respectively.

⇒ 21b + 19b = 40000

⇒ b = 1000

Number of students who choose Computer = 19b = 19000

Let a number of boys students who choose Maths be M.

The number of boys students who choose Computer = M – M × 50/100 = 50M/100 = M/2

Then,

⇒ 18000 = M + M/2

⇒ M = 12000

Number of boys students who choose Maths = 12000

Number of boys students who choose Computer = M/2 = 6000

Number of girls students who choose Computer = 19000 – 6000 = 13000

Required percentage = (22000 – 13000)/22000 × 100%

⇒ 40.90% ≈ 41%

The ratio of their profit = (A's investment × time) : (B’s investment × time)

⇒ (4 × 4 + (4 – 4/5) × 10) ∶ (5 × 6 + (5 – 5/4) × 8)

⇒ (16 + 32) ∶ (30 + 30) = 48 ∶ 60 = 4 ∶ 5

A's share in profit = 3600 × 4/9 = 1600

∴ A's share in the profit is Rs.1600.

Given:

Average marks = 65

Total number of students = 5

One student right marks = 37

Misread marks = 73

Formula Used:

Average = Sum of observations/Number of observations

Calculation:

According to the formula used

Total marks = 65 × 5 = 325

⇒ Difference between misread marks and right marks = 73 – 37 = 36

Now, 36 marks more are added to total

⇒ 325 – 36 = 289

GIVEN:

Ratio of sides of a cuboid is 6 : 4 : 5 and its volume is 3240 cm³.

Total surface area of the cuboid is 156 cm² more than the total surface area of a cube.

FORMULA USED:

Surface area of cuboid = 2 × [lb + bh + hl]

Surface area of cube = 6 × (Side)²

Volume of cuboid = lbh

CALCULATION:

Suppose the sides of the cuboid are 6a, 4a, and 5a respectively.

According to the question,

6a × 4a × 5a = 3240

⇒ 120a³ = 3240

⇒ a³ = 27

⇒ a = 3

Side of cuboid = 18, 12 and 15 cm

Suppose the side of cube = x cm

Now,

2 × (18 × 12 + 12 × 15 + 18 × 15) = 6x² + 156

⇒ 6x² = 2 × (216 + 180 + 270) – 156

⇒ 6x² = 1332 – 156

⇒ 6x² = 1176

⇒ x² = 196

⇒ x = 14 cm

∴ Side of cube = 14 cm

Calculation:

Let’s assume that the two qualities of tea need to be mixed in the ratio x : y.

I.e., x kgs of tea worth Rs. 27 should be mixed with y kgs of tea worth Rs. 31.

Total cost price of the mixture = 27x + 31y

Now, when the mixture is sold at Rs. 36 per kg, total selling price = 36x + 36y

Gain= 25%

⇒ (9x + 5y) × 4 = 27x + 31y

⇒ 36x + 20y = 27x + 31y

⇒ 9x = 11y

⇒ x/y = 11/9

Hence, the correct answer is 11 : 9.

The correct answer is option E.
Given:
An amount of Rs. 2,430 is divided between A, B and C such that if their shares be reduced by 5, 10 and
15 rupees respectively, the remainders shall be in the ratio of 3 ∶ 4 ∶ 5.
Calculation:
⇒ Total shares reduced of all three persons = 5 + 10 + 15 = 30
The remaining amount = 2430 - 30 = 2400
⇒ Now, ratio of A : B : C = 3 : 4 : 5

So, Share of B's in 2400 = = 800

⇒ B's earlier reduced share = Rs. 10

However, B's share = 800 + 10 = Rs. 810

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

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

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

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

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

Rounding off the values in the expression we get

√841 + √361 - √324 = ? × √225

⇒ 29 + 19 - 18 = ? × 15

⇒ 30 = ? × 15

⇒ ? = 30/15 = 2

Concept used:

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

494 - 15 × 8 + 4³ - 77 = ?²

⇒ 494 - 120 + 64 - 77 = ?²

⇒ 494 -120 - 13 = ?²

⇒ 494 - 133 = ?²

⇒ 361 = ?²

∴ The value of ? is 19.

Let the length of train be x meters

Relative speed of the train with respect to man = 90 + 35 = 125 km/hr

⇒ 125 × 5/18 = 34.72 m/s

Length of the train (x) = 34.72 × 15 = 520.8 meters

Let l be the length of platform

⇒ (520.8 + l) /45 = 90 × 5/18

⇒ l = 1125 - 520.8 = 604.2 meters

∴ Length of the train is 520.8 meters and length of the platform is 604.2 meters

From the given data,

⇒ x² – 15x + 56 = 0

⇒ x² - 7x - 8x + 56 = 0

⇒ x (x - 7) - 8 (x - 7) = 0

⇒ (x - 7)(x - 8) = 0

∴ x = 7 and x = 8

Also given that y² – 4y - 5 = 0

⇒ y² + y - 5y - 5 = 0

⇒ y (y + 1) - 5 (y + 1) = 0

⇒ (y - 5)(y + 1) = 0

∴ y = 5 and y = - 1

When x = 7 and y = 5, then x > y

When x = 7 and y = - 1, then x > y

When x = 8 and y = 5, then x > y

x = 8 and y = - 1, then x > y

Follow the 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

120 ÷ 5 × (3)³ ÷ 9 = (?) ÷ [8 ÷ (?)]

⇒ 120 ÷ 5 × (3)3 ÷ 9 = (?) ÷ (8/?)

⇒ 120 ÷ 5 × (3)³ ÷ 9 = (?) × (?)/8

⇒ (?)² = 24 × 3 × 8

7985 ÷ 115 × 22 = ? × 8

⇒ 69.435 × 22/8 = ?

Given:

The volume of the first one is thrice of the volume of the second one

The ratio of their height is 5 ∶ 2

Formula used:

The volume of a right circular cylinder = πr²h

Calculation:

Let the height of the first one be 5h and the height of the second one be 2h and also their radii be r₁ and r₂ respectively.

The volume of the first one is thrice of the volume of the second

So, the volume of the first one = πr₁²(5h)

The volume of the second one = πr₂²(2h)

According to the question, πr₁²(5h) = 3 × πr₂²(2h)

⇒ (r₁)²/(r₂)² = (6/5)

⇒ (r₁)/(r₂) = (√6/√5)

⇒ r₁ ∶ r₂ = √6 ∶ √5

∴ The ratio of their radii is √6 ∶ √5.

From the given table, Number of books published in Mathematics = 1840
Number of books published in Biology = 1480
ratio of number of books published in Mathematics and Biology = 1840 : 1480 = 46 : 37

From the given table, Number of Books Published in Economics subject = 1520
Number of Books Published in Chemistry Subject = 1900
required percentage = (1520/1900) * 100 = 80%

number of Books Published in Physics = 1260

number of Books Published in Biology = 1480

average number of Books Published in Physics & Biology Subject together = 1260 + 1480/2 = 1370

From given Table, number of books Published in chemistry = 1900
sold books in Chemistry =( 70/100) * 1900 = 1330

Calculation:

The series follows the following pattern

⇒ 120 - 18 = 102

⇒ 102 - 16 = 86

⇒ 86 - 14 = 72

⇒ 72 - 12 = 60

⇒ 60 - 10 = 50

∴ The wrong term is 84

Given series:

144, 1197, 196, 3375, 256

Calculation:

The series follows the following pattern:

12² = 144

13³ = 2197

14² = 196

15³ = 3375

16² = 256

The wrong term is 1197

Hence the wrong term in the series is 1197.

Given series:
5, 11, 19, 29, 40, 55
Calculation :
The series follows the following pattern:
⇒ 2 + 2² - 1 = 5
⇒ 3 + 3² - 1 = 11
⇒ 4 + 4² - 1 = 19
⇒ 5 + 5² - 1 = 29
⇒ 6 + 6² - 1 = 41
⇒ 7 + 7² - 1 = 55
The wrong term is 40
Hence the wrong term in the series is 40.

The logic followed here is as follows:

⇒ 1³ = 1

⇒ 2³ = 8

⇒ 3³ = 27

⇒ 4³ = 64

⇒ 5³ = 125

⇒ 6³ = 216

So, the wrong term is 344

Hence, the correct answer is 344.

Given:
√400 ÷ 3.87 + (4.96)² = ?
Concept used:
Follow the BODMAS rule according to the table given below:

Calculation:

√400 ÷ 3.87 + (4.96)² = ?

⇒ 20 ÷ 4 + 52 = ?

⇒ 20/4 + 25 = ?

⇒ 5 + 25 = ?

⇒ ? = 30

∴ The value of ? is 30.

Concept:
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.

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

(6318 – 6168) × 6 = x²

150 × 6 = x²

⇒ 900 = x²

⇒ X = 30

Given:

A steamer can move 36 km in downstream = 6 hr

A steamer can move 36 km in upstream = 10 hr

Formula used:

Speed of the stream = (Speed in downstream - speed in upstream)/2

Calculation:

Speed in downstream = 36/6 = 6 km/h

Speed in upstream = 36/10 = 3.6 km/h

Speed of stream = (6 - 3.6)/2 = 1.2 km/h

∴ The speed of the stream is 1.2 km/h.