CUET PG MCA Quantitative Aptitude Previous Year Questions (PYQs)
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A sum of money doubles itself on simple interest in 10 years.
Find the rate of interest per annum.
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Solution
Let the principal be \( P \).
Since the money doubles in 10 years, the total amount becomes \( 2P \).So, Simple Interest (SI) = \( 2P - P = P \)
Use the formula:
$$\text{SI} = \frac{P \cdot R \cdot T}{100}$$ Substituting values: $$P = \frac{P \cdot R \cdot 10}{100}$$ Cancel \( P \) on both sides: $$1 = \frac{R \cdot 10}{100} \Rightarrow R = \frac{100}{10} = \boxed{10\%}$$
Final Answer:
$$\boxed{10\% \text{ per annum}}$$
In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If
she/he attempts all 60 questions and secures 130 marks, the number of questions she/he attempts wrongly, are?
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Solution
Let the number of correct answers be \(x\) and wrong answers be \(y\).
Total questions: \[ x + y = 60 \tag{1} \] Marks rule: \(4\) marks for correct and \(-1\) mark for wrong, total \(130\). \[ 4x - y = 130 \tag{2} \] From (1): \(y = 60 - x\). Substituting in (2): \[ 4x - (60 - x) = 130 \;\;\Rightarrow\;\; 5x - 60 = 130 \;\;\Rightarrow\;\; x = 38 \] Hence, \[ y = 60 - 38 = 22 \]
Total questions: \[ x + y = 60 \tag{1} \] Marks rule: \(4\) marks for correct and \(-1\) mark for wrong, total \(130\). \[ 4x - y = 130 \tag{2} \] From (1): \(y = 60 - x\). Substituting in (2): \[ 4x - (60 - x) = 130 \;\;\Rightarrow\;\; 5x - 60 = 130 \;\;\Rightarrow\;\; x = 38 \] Hence, \[ y = 60 - 38 = 22 \]
The area of a rhombus is 120 $cm^2$ and length of its one diagonal
in 24 cm. Find the perimeter of the rhombus (in cm)
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Solution
The monthly income and expenditure of a person were Rs.10,000
and Rs. 6,000 respectively. Next year, his income increased by
15% and his expenditure increased by 8%. Then the percentage
increase in his savings is:
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Solution
In a class, 4/5 of the students are boys and rest are girls. If 2/5 of the boys and 1/4 of girls are absent, what part of the total number of students is present?
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Solution
Solution
Given: 4/5 of the students are boys, and the rest are girls. 2/5 of boys and 1/4 of girls are absent.
Step-by-Step Solution
- Total students =
N - Boys:
(4/5) × N = 4N/5 - Girls:
(1/5) × N = N/5 - Boys absent:
(2/5) × (4N/5) = 8N/25 - Girls absent:
(1/4) × (N/5) = N/20 - Boys present:
(4N/5 - 8N/25 = 12N/25) - Girls present:
(N/5 - N/20 = 3N/20) - Total present =
(12N/25 + 3N/20)
Final Calculation
LCM of 25 and 20 is 100:
(12N/25 = 48N/100), (3N/20 = 15N/100)
Total present = (48N/100 + 15N/100 = 63N/100)
Answer: 63/100
If 2 cats catch 2 mice in 2 minutes, how many cats are needed to catch 18 mice in 18 minutes?
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Solution
By the formula
$\frac{M_1 H_1 D_1}{W_1}$=$\frac{M_2 H_2 D_2}{W_2}$ ----(i)
Here
$M_1=2, H_1=2, W_1=2$
$M_2=?, H_2=18, W_2=18$
Using (i)
$\frac{2 \times 2 \times 1}{2}$=$\frac{M_2 \times 18 }{18}$
$M_2=2$
Hence, 2 cats are needed to catch 18 mice.
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