RRB NTPC CBT 2 Level 5 June-12-2022 Shift 1 Exam Previous Question Paper with Solutions

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106) A sum of money at certain rate of interest when compounded annually becomes ₹625 in 3 years and ₹675 in 4 years. What is the rate of Interest per annum?

A) 8%
B) 4%
C) 7%
D) 6%

View Answer

A) 8%

Explanation:We are given:
* Amount after 3 years = ₹625
* Amount after 4 years = ₹675
The interest for the 4th year alone = ₹675 – ₹625 = ₹50
This is because compound interest adds interest on interest, so the 4th year’s interest is calculated on the amount at the end of 3rd year (₹625).
Step 1: Use formula for Compound Interest:
Let the rate be $R\%$ per annum.
We know:
$\text{CI for 4th year} = \text{Amount after 3 years} × \frac{R}{100}$
So:
$50 = 625 × \frac{R}{100}$
$\frac{R}{100} = \frac{50}{625} = \frac{2}{25} ⇒ R = \frac{2}{25} × 100 = 8\%$
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✅ Final Answer: 1. 8%

107) How many technology innovation platforms have been launched by the Central Government for development of technologies for globally competitive manufacturing in India in July 2021?

A) Six
B) Five
C) Eight
D) Seven

View Answer

A) Six

108) If 32% of 22.5 $-\frac23\times\sqrt[3]{512}\times\sqrt{81}$ =y,then the value of y is:

A) -40.2
B) -41.2
C) -40.8
D) -41.8

View Answer

C) -40.8

Explanation:We are given:
$32\% \text{ of } 22.5 - \frac{2}{3} × \sqrt[3]{512} × \sqrt{81} = y$
Step 1: Calculate each part
(i) $32\% \text{ of } 22.5$
$= \frac{32}{100} × 22.5 = 0.32 × 22.5 = 7.2$
(ii) $\sqrt[3]{512}$
$\sqrt[3]{512} = 8 \quad (\text{since } 8^3 = 512)$
(iii) $\sqrt{81}$
$\sqrt{81} = 9$
Now compute:
$\frac{2}{3} × 8 × 9 = \frac{2}{3} × 72 = 48$
Step 2: Plug values into original expression:
7.2 – 48 = -40.8
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✅ Final Answer: 3. -40.8

109) Evaluate $\frac{\cos e^2\left(45^\circ+\theta\right)+\cos^2\left(45^\circ-\theta\right)}{\cos ec^230^\circ\sin^245^\circ-sec^260^\circ}$

A) $-\frac12$
B) $-\frac16$
C) $\frac16$
D) $\frac12$

View Answer

A) $-\frac12$

Explanation:We are asked to evaluate:
$\frac{\cos^2(45^\circ+\theta) + \cos^2(45^\circ - \theta)}{\csc^2 30^\circ \sin^2 45^\circ - \sec^2 60^\circ}$
—
Step 1: Simplify the Numerator
Use the identity:
$\cos^2 A + \cos^2 B = 1 + \cos(A+B) \cos(A-B)$
Alternatively, for specific symmetric cases like this:
$\cos^2(45^\circ + \theta) + \cos^2(45^\circ - \theta)$
Use the identity:
$\cos^2 x = \frac{1 + \cos 2x}{2}$
So,
$\cos^2(45^\circ + \theta) = \frac{1 + \cos(90^\circ + 2\theta)}{2} = \frac{1 - \sin(2\theta)}{2}$
$\cos^2(45^\circ - \theta) = \frac{1 + \cos(90^\circ - 2\theta)}{2} = \frac{1 + \sin(2\theta)}{2}$
Add them:
$\frac{1 - \sin(2\theta)}{2} + \frac{1 + \sin(2\theta)}{2} = \frac{1 + 1}{2} = 1$
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Step 2: Simplify the Denominator
$\csc^2 30^\circ = \left(\frac{1}{\sin 30^\circ}\right)^2 = \left(\frac{1}{1/2}\right)^2 = 4$
$\sin^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$
$\sec^2 60^\circ = \left(\frac{1}{\cos 60^\circ}\right)^2 = \left(\frac{1}{1/2}\right)^2 = 4$
Now compute the denominator:
$4 × \frac{1}{2} - 4 = 2 - 4 = -2$
—
Step 3: Final Evaluation
$\frac{\text{Numerator}}{\text{Denominator}} = \frac{1}{-2} = -\frac{1}{2}$
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✅ Final Answer: 1. $-\frac{1}{2}$