Numerical Ability Test - 5 - Bank Exams MCQ

Concept used :

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

⇒ 38% of 150 + 18% of 50 = ? + 10

⇒ 38 × 150 /100 + 18 × 50 / 100 = ? + 10

⇒ 57 + 9 = ? + 10

⇒ 66 = ? + 10

⇒ ? = 66 - 10

⇒ ? = 56

Hence 56 is the correct answer.

Given:

60% of 320 + 50% of 190 - 57% of 350 = ?

Concept Used:

​

Calculation:

60% of 320 + 50% of 190 - 57% of 350 = ?

⇒ 192 + 95 - 199.5

⇒ ? = 87.5

∴ The value of (?) is 87.5

Area of the circle = π × r²

ATQ,

⇒ 15400 = π × r × r

⇒ r = 70 cm.

radius including the path width = r + 7 or 77 cm

Area of the path = area of the park including path - area of the park without path

⇒ A = π × 77 × 77 - 15400

⇒ A = 3234 cm²

Cost of making path = 3234 × 20

In Simple interest,

Principal, P = Rs. 5700

Rate, R = 5%

Time, T = 2 years

∴ Simple interest earned,

⇒ (5700 × 5 × 2)/100

⇒ Rs. 570

In compound interest,

Principal, P = Rs. 5700

Rate, R = 5%

No. of years, n = 2

∴ Compound interest earned,

⇒ P (1 + R/100)ⁿ - P

⇒ 5700 × (1 + 5/100)² - 5700 = 6284.25 - 5700 = Rs. 584.25

Given expression:

⇒ 55% of 200 + ? = 22% of 400 + 6% of 460

⇒

⇒ 110 + ? = 88 + 27.6

⇒ ? = 115.6 - 110

⇒ ? = 5.6

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

324 ÷ 9 × 8 – 48 = ?

⇒ 36 × 8 – 48 = ?

⇒ 288 – 48 = ?

∴ ? = 240

Given:
336 ÷ 24 × ∛343 + (7)² = ?
Concept:
Follow BODMAS rule to solve this question, as per the order given below,

Calculation:
336 ÷ 24 × ∛343 + (7)² = ?
⇒ 14 × 7 + 49 = ?
⇒ 98 + 49 = ?
∴ ? = 147

First of all make table --

Total turn over of jeera biscuit for all four months = ( 200 + 220 + 240 + 260 ) × 50
= 920 × 50 = 46000
Total turn over of cake biscuit for all four months = ( 160 + 176 + 192 + 208 ) × 100
= 736 × 100 = 73600
∴ Required sum = 73600 + 46000 = 119600

First of all make table --

Total sale price for month of July = ( 200 × 50 + 150 × 60 + 170 × 70 + 160 × 100 ) = 10000 + 9000 + 11900 + 16000 = 46900

total sale price for month of October = ( 260 × 50 + 195 × 60 + 221 × 70 + 208 × 100 ) = 13000 + 11700 + 15470 + 20800 = 60970

∴ required ratio = 60970 / 46900 = 1.3

First of all make table --

total selling price of sweet biscuit = 690 × 60 = 41400
total selling price of cake biscuit = 736 × 100 = 73600
∴ required % = ( 73600 – 41400 ) × 100 / 41400 = 77.78 %

x =

⇒

⇒ x = 19

y² = 361

⇒ y =

⇒ y = (+19, -19)

x² - 34x + 288 = 0

⇒ x² - 16x - 18x + 288 = 0

⇒ (x - 16) (x - 18) = 0

⇒ x = 16 or x = 18

y² - 35y + 306 = 0

⇒ y² - 17y - 18y + 306 = 0

⇒ (y - 17) (y - 18) = 0

⇒ y = 17 or y = 18

Hence, Relation between x and y cannot be established.

(I) m² – 7m + 10 = 0

⇒ m² – 5m – 2m + 10 = 0

⇒ m(m – 5) – 2(m – 5) = 0

⇒ (m – 5)(m – 2) = 0

⇒ m = (5, 2)

(II) n² + 8n + 15 = 0

⇒ n² + 5n + 3n + 15 = 0

⇒ n(n + 5) + 3(n + 5) = 0

⇒ (n + 5)(n + 3) = 0

⇒ n = (-5, -3)

Hence, m > n.

Quantity of milk in Drum A = 3 litre

Quantity of water in Drum A = 5 litres

Total Quantity = 8 litres

Quantity transferred to Drum B = 60% of 8

= 0.6 × 8

= 4.8 litres

It contains 1.8 litres of milk and 3 litres of water

Quantity of milk retained in Drum A = 1.2 litres

Quantity of extra water added = 2 litres

Total liquid in Drum B = 6.8 litres.

Quantity of milk in Drum B = 1.8 litres.

⇒ Ratio of milk in drum A to B = 1.2 ∶ 1.8

= 2 : 3

I. x² – 7x + 12 = 0

⇒ x² – 4x – 3x + 12 = 0

⇒ x(x – 4) – 3(x – 4) = 0

⇒ (x – 3) (x – 4) = 0

⇒ x = + 3 OR x = + 4

II. y² – 9y + 20 = 0

⇒ y² – 5y – 4y + 20 = 0

⇒ y(y – 5) – 4(y – 5) = 0

⇒ (y – 4) (y – 5) = 0

⇒ y = + 4 OR y = + 5

So, when x = 3 , x < y for y = 4 and x

And when x = 4 , x = y for y = 4 and x

∴ x ≤ y.

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

⇒ (74 × 4 – 66) ÷ 5 = 89 + 37 - ?

⇒ (296 – 66) ÷ 5 = 89 + 37 – ?

⇒ 230 ÷ 5 = 126 – ?

⇒ ? = 126 – 46

⇒ ? = 80

∴ The value of ? is 80

Concept Used:

Calculation:

(60% of 700) ÷ 6 + 5 = ?

⇒ [(60/100) × 700] ÷ 6 + 5 = ?

⇒ (60 × 7) ÷ 6 + 5 = ?

⇒ 420 ÷ 6 + 5 = ?

⇒ 70 + 5 = ?

∴ ? = 75

Given

Rajan’s age = Samara’s age + 5

Ratio of their ages after 5 years = 7 ∶ 6

Formula used

Ratio formula, a ∶ b = a/b

Calculation

Let the age of Rajan = x

And let the age of Samara = y

According to question

⇒ x = y + 5

⇒ y = x – 5

After 5 years

⇒ (x + 5)/(y + 5) = 7/6

⇒ 6x + 30 = 7y + 35

⇒ 6x – 7y = 5

⇒ 6x – 7(x – 5) = 5

⇒ x = 30 years

Rajan’s age = 30 years

Rajan’s age 10 years ago = 30 – 10 = 20 years

∴ Rajan’s age 10 years ago is 20 years.

Nani’s monthly income in all the years together = 17 + 20 + 22 + 23 + 22 + 37 = 141
Xavier’s monthly income in the year 2014 = 27
∴ Required difference = 141 – 27 = 114 × 1000 = 114000 = 1.14 lakh

Required ratio = (21 + 33) ∶ (18 + 30) ∶ (17 + 37) = 54 ∶ 48 ∶ 54 = 9 ∶ 8 ∶ 9

The difference between Xavier’s and Nitin’s monthly income in 2010 = 21 – 18 = 3
The difference between Xavier’s and Nitin’s monthly income in 2011 = 23 – 20 = 3
The difference between Xavier’s and Nitin’s monthly income in 2012 = 27 – 22 = 5
The difference between Xavier’s and Nitin’s monthly income in 2013 = 32 – 23 = 9
The difference between Xavier’s and Nitin’s monthly income in 2014 = 35 – 27 = 8
The difference between Xavier’s and Nitin’s monthly income in 2015 = 33 – 30 = 3
Hence, 2014 was the year in which the difference between Xavier’s and Nitin’s monthly income the second-highest

Monthly income of Nani in the year 2015 = 37
Monthly income of Nani in the year 2014 = 22
∴ Required % increase = [(37 – 22)/22] × 100 = 68%

Monthly income of xavier in 2011 = 20
Monthly income of Nitin in 2013 = 32
∴ Required Percentage = [(32 - 20)/32] × 100 = 37.5%

Calculation:
According to the question,
Ram can dig a ditch alone in 30 days and together with Mohan they can dig it
in 12 days,
Then Mohan's digging speed = 1/30 - 1/12 = 1/20 (ditches per day).
Therefore, Mohan can dig the whole ditch alone in 20 days.
So, the answer is (C).

Given:

The capacity of vessel = 120 litres

Calculation:
Let the quantity should be replaced be x litres

Then, Initial quantity of Spirit = [7/(7 + 5)] × 120 litres

⇒ (7/12) × 120 litres

⇒ 70 litres

and, Initial quantity of water = [5/(7 + 5)] × 120 litres

⇒ (5/12) × 120 litres

⇒ 50 litres

Now, [70 - (7/12) × x + x]/[50 - (5/12) × x] = (5/3)

⇒ (840 - 7x + 12x)/(600 - 5x) = (5/3)

⇒ (840 + 5x) × 3 = (600 - 5x) × 5

⇒ 2520 + 15x = 3000 - 25x

⇒ 40x = 480

⇒ x = 12 litres

The fraction of spirit in the initial mixture = 7/12

Fraction of spirit present in spirit alone = 1

Resultant fraction of spirit in final mixture = 5/8

By the process of alligation:

Ratio in which they were mixed = 9 ∶ 1

The total quantity of mixture remains the same = 120

So, 10 units = 120 L

Then amount of spirit added = 1 units = 12 L

∴ 12 litres of the mixture should be replaced by spirit to make the ratio 5 : 3.

Given:

4.99 × 15.01 + 29.99 × 8 = ? × 5.01

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

4.99 × 15.01 + 29.99 × 8 = ? × 5.01

⇒ 5 × 15 + 30 × 8 = ? × 5

⇒ 75 + 240 = ? × 5

⇒ ? = 315/5

⇒ ? = 63

∴ The value of ? will be 63.

Given:
10.10 × √(3.98) + 24.99% of 28.02 = √?
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 used:
Follow the BODMAS rule according to the table given below:

Calculation:
10.10 × √(3.98) + 24.99% of 28.02 = √?
⇒ 10 × √4 + 25% of 28 = √?
⇒ 10 × 2 + (25/100) × 28 = √?
⇒ 20 + 7 = √?
⇒ √? = 27
⇒ ? = 27²
⇒ ? = 729
∴ 729 should come in place of the question mark (?).

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

Calculation:

25.01% of 976.22 × 4.96 = ?

25/100 × 976 × 5 = ?

⇒ 1/4 × 976 × 5 =?

⇒ 244 × 5 = ?

⇒ ? = 1220

∴ The required answer = 1220

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

Calculation:

[(3.96)% of √99.99] × √25 = ?

Approximating the values,

(4% of √100) × √25 = ?

[4% of 10] × 5 = ?

[(4/100) × 10] × 5 = ?

0.4 × 5 = ?

∴ ? = 2

Given:

Single Discount = 25%

Successive Discount = 10% and 20%

Difference = 1800

Calculation:

Let the bill amount be Rs. 100x.

Case 1:

⇒ 25% discount to the 100x = 25x discount

⇒ Price after discount = 100x - 25x = 75x

Case 2:

⇒ Discount of 10% to the 100x = 10x

⇒ Price after discount = 100x - 10x = 90x

⇒ Second Discount of 20% to the 90x = 18x

⇒ Price after second discount = 90x - 18x = 72x

Difference between case 1 and case 2 is given as 1800.

⇒ 75x - 72x = 1800

⇒ 3x = 1800

⇒ x = 1800/3

⇒ x = 600

⇒100 x = Rs. 60,000

∴ The bill amount is of Rs. 60,000.

Hint

Successive discount can be solved quickly by:

Discount = x + y – (xy/100)

⇒ -10 – 20 - [(-10 – 20)/100]

⇒ -30 + 2

⇒ -28%