RRB NTPC CBT 2 Level 6 May-2-2022 Shift 1 Exam Previous Question Paper with Solutions

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66) Ram gave a discount of 10% on the marked price of an item and still gained 12.5%. How much would have Ram gained, if he sold the item at the marked price?

A) 27.5%
B) 25%
C) 20%
D) 22.5%

View Answer

B) 25%

Explanation:To solve this problem, we need to calculate the profit Ram would have made if he sold the item at the marked price, given that he gave a 10% discount and still gained 12.5%.
Step-by-Step Solution:
Let the marked price of the item be ( M ), and let the cost price of the item be ( C ).
Step 1: Calculate the selling price with the discount
– Ram gave a 10% discount on the marked price, so the selling price after the discount is:
$\text{Selling Price} = M - 10\% \text{ of } M = M - \frac{10}{100} × M = 0.9M$
Step 2: Use the gain percentage to relate the cost price and selling price
– Ram gained 12.5% on the cost price. The selling price is therefore 112.5% of the cost price:
$\text{Selling Price} = C + 12.5\% \text{ of } C = C + \frac{12.5}{100} × C = 1.125C$
We know that the selling price after the discount is ( 0.9M ), and it is also equal to ( 1.125C ) (from the gain). Therefore, we can equate the two expressions for the selling price:
0.9M = 1.125C
Step 3: Find the relationship between the marked price and the cost price
Solving for ( M ) in terms of ( C ):
$M = \frac{1.125C}{0.9} = 1.25C$
Step 4: Calculate the gain if Ram sold at the marked price
If Ram had sold the item at the marked price ( M ), the selling price would have been ( M = 1.25C ). The profit would have been:
Profit = M – C = 1.25C – C = 0.25C
The gain percentage is the profit as a percentage of the cost price:
$\text{Gain Percentage} = \frac{0.25C}{C} × 100 = 25\%$

67) Find the value of the expression:
$1-\frac{(4÷5-1\times3+2)\times8}{3^2\times8-4\times2}$

A) $\frac{39}{40}$
B) $\frac{41}{40}$
C) 0
D) $\frac58$

View Answer

B) $\frac{41}{40}$

Explanation:Let’s evaluate the expression step by step:
$1 - \frac{(4 ÷ 5 - 1 × 3 + 2) × 8}{3^2 × 8 - 4 × 2}$
Step 1: Simplify the numerator
The numerator is:
(4 ÷ 5 – 1 × 3 + 2) × 8
First, solve the operations inside the parentheses:
»1. ( 4 ÷ 5 = 0.8 )
»2. ( 1 × 3 = 3 )
Now, substitute these values into the expression:
0.8 – 3 + 2 = -0.2
So, the numerator becomes:
(-0.2) × 8 = -1.6
Step 2: Simplify the denominator
The denominator is:
$3^2 × 8 - 4 × 2$
First, simplify each part:
»1. ( 3^2 = 9 )
»2. ( 9 × 8 = 72 )
»3. ( 4 × 2 = 8 )
Now, subtract:
72 – 8 = 64
Step 3: Combine the numerator and denominator
Now we substitute the simplified numerator and denominator back into the expression:
$1 - \frac{-1.6}{64}$
Simplify the fraction:
$\frac{-1.6}{64} = \frac{-16}{640} = \frac{-1}{40}$
So, the expression becomes:
$1 - \left( \frac{-1}{40} \right) = 1 + \frac{1}{40} = \frac{40}{40} + \frac{1}{40} = \frac{41}{40}$
Final Answer:
The value of the expression is ( $\frac{41}{40}$ ).

68) A consumer’s optimal bundle is located at the point of tangency between the budget line and __________

A) indifference curve
B) law of demand
C) income effect
4 monotonicity

View Answer

A) indifference curve
4 monotonicity

69) Ramesh ranked 16th from the top and 21st from the bottom in his class in an examination. If only 21 students passed the exam, how many students in his class failed?

A) 19
B) 17
C) 15
D) 21

View Answer

C) 15

Explanation:We are given the following information:
– Ramesh ranked 16th from the top.
– Ramesh ranked 21st from the bottom.
– 21 students passed the exam.
Step 1: Find the total number of students in the class
Let the total number of students in the class be ( N ).
From the given information, we know that Ramesh’s rank from the top is 16th, and his rank from the bottom is 21st. This means that there are 15 students above him and 20 students below him (since Ramesh is counted in both ranks).
Thus, the total number of students is:
N = 15 (students above Ramesh) + 1 (Ramesh himself) + 20 (students below Ramesh) = 36
Step 2: Find the number of students who failed
We are told that 21 students passed the exam. Since the total number of students is 36, the number of students who failed is:
Number of students who failed} = 36 – 21 = 15

70) Which ceremonial folk dance of Puducherry is related to the Hindu epic Ramayana, performed at the Villianur temple in Puducherry?

A) Garadi
B) Hulivesha
C) Dalkhai
D) Tippani

View Answer

A) Garadi

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