CSAT MCQ Quiz - Objective Question with Answer for CSAT - Download Free PDF
Last updated on Jun 5, 2025
Latest CSAT MCQ Objective Questions
CSAT Question 1:
The difference between any two natural numbers is 10. What can be said about the natural numbers which are divisible by 5 and lie between these two numbers?
Answer (Detailed Solution Below)
CSAT Question 1 Detailed Solution
The Correct answer is Option 3
Key Points
If the numbers are 1 and 11, there are two numbers between them that are divisible by 5 (i.e. 5 and 10).
However, if we consider the numbers 5 and 15, there is only one number between them that is divisible by 5 (i.e. 10).
Hence, there can be one or two such numbers in the given range, depending on the range we choose.
Hence the Correct answer is Option 3.
CSAT Question 2:
Let x be a real number between 0 and 1. Which of the following statements is/are correct?
I. x2 > x3
II. x > √x.
Select the correct answer using the code given below:
Answer (Detailed Solution Below)
CSAT Question 2 Detailed Solution
The Correct answer is Option 1
Key Points
Given that: 0 < x < 1
Statement I: x2 > x3
This statement is correct.
For example, let value of x be 0.5.
x2 = 0.52 = 0.25
x3 = 0.53 = 0.125
Statement II: x > √x
Let’s check with the help of an example.
If x = 0.25
Then, √x = √0.25 = 0.5
So, it’s evident that this statement is not correct.
Hence, option (a) is the correct answer.
CSAT Question 3:
The average of three numbers p, q and r is k. p is as much more than the average as q is less than the average. What is the value of r?
Answer (Detailed Solution Below)
CSAT Question 3 Detailed Solution
The Correct answer is Option 1
Key Points
Let p is x more than the average, and q is x less than the average.
So, p = k + x, q = k – x
We also know that, the average of p, q and r is k.
So, (p + q + r)/3 = k
Or {(k + x) + (k - x) + r} / 3 = k
Or 2k + r = 3k
Or, r = k
Hence the Correct answer is Option 1.
CSAT Question 4:
Consider a set of 11 numbers:
Value-I = Minimum value of the average of the numbers of the set when they are consecutive integers -5.
Value-II = Minimum value of the product of the numbers of the set when they are consecutive non-negative integers.
Which one of the following is correct?
Answer (Detailed Solution Below)
CSAT Question 4 Detailed Solution
The Correct answer is Option 3
Key Points
Total numbers = 11
For Value I:
Minimum value of average is possible when the numbers being considered are the smallest possible.
Smallest 11 consecutive numbers ≥ -5 are: -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5
Their average, Value I = 0
Value II:
Smallest non-negative integer is 0.
So, the minimum possible value of the product of consecutive non-negative integers, Value II = 0
So, Value I = Value II
Hence the Correct answer is Option 3.
CSAT Question 5:
Let p + q = 10, where p, q are integers.
Value-I = Maximum value of p × q when p, q are positive integers.
Value-II = Maximum value of p × q when p ≥ -6, q ≥ -4.
Which one of the following is correct?
Answer (Detailed Solution Below)
CSAT Question 5 Detailed Solution
The Correct answer is Option 3
Key Points
Given that p + q = 10, where p and q are integers.
When sum of two numbers is constant, their multiple is the maximum when their values are as close as possible.
So, Value I = 5 × 5 = 25
For p × q to be maximum, they both must be negative or both positive.
The maximum value of p × q, when both numbers are negative = (-6) × (-4) = 24
The maximum value of p × q, when both numbers are positive = 5 × 5 = 25
So, Value II = 25
Hence, Value I = Value II
Hence the Correct answer is Option 3.
Top CSAT MCQ Objective Questions
सूची - I में चार संज्ञा शब्द एवं सूची - II में संबंधित संज्ञा के प्रकार दिए गए हैं । सूची - I का मिलान, सूची - II से कीजिए तथा नीचे दिए गए कूट से सही उत्तर चुनिए :
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सूची - I |
सूची - II |
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(a) |
कक्ष |
(i) |
व्यक्तिवाचक संज्ञा |
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(b) |
कक्षा |
(ii) |
भाववाचक संज्ञा |
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(c) |
सजावट |
(iii) |
जातिवाचक संज्ञा |
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(d) |
महाभारत |
(iv) |
समूहवाचक संज्ञा |
Answer (Detailed Solution Below)
CSAT Question 6 Detailed Solution
Download Solution PDFसही विकल्प 2 है।Key Pointsसही मिलान होगा-
|
सूची - I |
सूची - II |
||
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(a) |
कक्ष |
(i) |
जातिवाचक संज्ञा |
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(b) |
कक्षा |
(ii) |
समूहवाचक संज्ञा |
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(c) |
सजावट |
(iii) |
भाववाचक संज्ञा |
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(d) |
महाभारत |
(iv) |
व्यक्तिवाचक संज्ञा |
सूची - I में चार शब्द दिए गए हैं तथा सूची - II में स्रोत के आधार पर शब्द भेद दिए गए हैं। सूची - I का मिलान, सूची - II से कीजिए तथा नीचे दिए गए कूट से सही उत्तर चुनिए :
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सूची - I |
सूची - II |
||
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(a) |
लता |
(i) |
तद्भव शब्द |
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(b) |
पत्ता |
(ii) |
तत्सम शब्द |
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(c) |
जूता |
(iii) |
विदेशज शब्द |
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(d) |
गुलाब |
(iv) |
देशज शब्द |
Answer (Detailed Solution Below)
CSAT Question 7 Detailed Solution
Download Solution PDFसही विकल्प 3 है।Key Pointsसही मिलान है-
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सूची - I |
सूची - II |
||
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(a) |
लता |
(i) |
तत्सम शब्द |
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(b) |
पत्ता |
(ii) |
तद्भव शब्द |
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(c) |
जूता |
(iii) |
देशज शब्द |
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(d) |
गुलाब |
(iv) |
विदेशज शब्द |
The remainder, when 1 + (1 × 2) + (1 × 2 × 3) +......+ (1 × 2 × 3 × .... × 500) is divided by 8, is
Answer (Detailed Solution Below)
CSAT Question 8 Detailed Solution
Download Solution PDFThe correct answer is Option 1
Key PointsTo find the remainder when
⇒ S=1+(1×2)+(1×2×3)+…+(1×2×3×…×500) is divided by 8, so we can analyze the terms in the series.
⇒ The nth term of the series is n! (n factorial). We need to find the sum of factorials from 1! to 500! and then find the remainder when this sum is divided by 8.
⇒ Calculating the factorials modulo 8:
- 1!=1≡1 mod 8
- 2!=2≡2mod8
- 3!=6≡6 mod 8
- 4!=24≡0 mod8
⇒ For n≥4 n! will always be divisible by 8 (since 4! and higher factorials include the factors 2 and 4). Thus, we only need to
consider the first three terms:
⇒ S≡1+2+6 mod 8
Calculating this:
⇒ S≡1+2+6=9 ≡1 mod 8
Therefore, the remainder when
S is divided by 8 is 1.
The ratio of present ages (in years) of X to Y is equal to the ratio of present ages (in years) of Y to Z. If the present age of Y is 15 years, then which of the following can be the sum of the ages (in years) of X, Y and Z?
Answer (Detailed Solution Below)
CSAT Question 9 Detailed Solution
Download Solution PDFThe Correct answer is Option 3.
Key PointsX/Y = Y/Z ( Equation )
⇒ XZ = Y2
⇒ Here, Y = 15 , Means XZ = 225
⇒ Factroize = 3 x 5 x 3 x 5
⇒ Case 1 = 3 x 3 x 5 x 5 = 9 x 25
Let Put X = 9 , Y= 15 , Z = 25 , Put in Above Equation
⇒ 9/15 = 15/25 , This is Perfect as per equation
⇒ Hence X + Y + Z = 9 + 15 + 25 = 49
Hence Correct answer is Option 3.
Which one of the following is the average of first five multiples of each of the numbers from 11, 12, 13,... 20 ?
Answer (Detailed Solution Below)
CSAT Question 10 Detailed Solution
Download Solution PDFThe correct answer is Option 4.
Key PointsCalculate the first five multiples of each number:
- For 11: 11,22,33,44,55
- For 12: 12,24,36,48,60
- For 13: 13,26,39,52,65
- For 14: 14,28,42,56,70
- For 15: 15,30,45,60,75
- For 16: 16,32,48,64,80
- For 17: 17,34,51,68,85
- For 18: 18,36,54,72,90
- For 19: 19,38,57,76,95
- For 20: 20,40,60,80,100
Sum the first five multiples for each number:
- Sum for 11: 11+22+33+44+55=165
- Sum for 12: 12+24+36+48+60=180
- Sum for 13: 13+26+39+52+65=195
- Sum for 14: 14+28+42+56+70=210
- Sum for 15: 15+30+45+60+75=225
- Sum for 16: 16+32+48+64+80=240
- Sum for 17: 17+34+51+68+85=255
- Sum for 18: 18+36+54+72+90=270
- Sum for 19: 19+38+57+76+95=285
- Sum for 20: 20+40+60+80+100=300
Total sum of all these sums:
Total Sum=165+180+195+210+225+240+255+270+285+300
Calculating this step-by-step:
- 165+180=345
- 345+195=540
- 540+210=750
- 750+225=975
- 975+240=1215
- 1215+255=1470
- 1470+270=1740
- 1740+285=2025
- 2025+300=2325
Calculate the average: There are 10 numbers (from 11 to 20), and each has 5 multiples, so we divide the total sum by
10×5=50
Average=232550=46.5
Thus, the average of the first five multiples of each of the numbers from 11 to 20 is 46.5
सूची - I में चार सामासिक शब्द तथा सूची - II में समास के नाम दिए गए हैं। सूची -I का मिलान, सूची - II से कीजिए तथा नीचे दिए गए कूट से सही उत्तर चुनिए :
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सूची - I |
सूची - II |
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(a) |
महोदधि |
(i) |
अव्ययीभाव समास |
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(b) |
वीणापाणि |
(ii) |
तत्पुरुष समास |
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(c) |
यथायोग्य |
(iii) |
कर्मधारय समास |
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(d) |
विद्यालय |
(iv) |
बहुव्रीहि समास |
Answer (Detailed Solution Below)
CSAT Question 11 Detailed Solution
Download Solution PDFसही विकल्प 2 है।Key Pointsसही मिलान होगा-
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सूची - I |
सूची - II |
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(a) |
महोदधि |
(i) |
कर्मधारय समास |
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(b) |
वीणापाणि |
(ii) |
बहुव्रीहि समास |
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(c) |
यथायोग्य |
(iii) |
अव्ययीभाव समास |
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(d) |
विद्यालय |
(iv) |
तत्पुरुष समास |
If \(\rm x+\frac{1}{x}=2\), then which one of the following is the value of \(\rm x^{32}+\frac{1}{x^{32}}\) ?
Answer (Detailed Solution Below)
CSAT Question 12 Detailed Solution
Download Solution PDFThe Correct answer is Option 4.
Key Points ⇒ ( x + 1/x )2 = 22
⇒ x2 + 1/x2 + 2 = 4
⇒ x2 + 1/x2 = 2
⇒ When We square of x2 + 1/x2 , we will get again 2
Hence Correct answer is Option 4.
Which of the following statements is/are correct?
1. The average of four numbers 10, 15, 20 and 25 is 17.5
2. If a, b and c are three different such that natural numbers a + b + c = abc, then the average of a, b and c is 3
Select the answer using the code given below:
Answer (Detailed Solution Below)
CSAT Question 13 Detailed Solution
Download Solution PDFThe correct answer is Option 1
Key PointsStatement 1: The average of four numbers 10, 15, 20, and 25 is 17.5.
⇒ To find the average, we use the formula: Average = Sum of numbers / Number of numbers
⇒ Calculating the sum: 10+15+20+25=70
⇒ Now, calculating the average: Average=70/4=17.5
Conclusion: Statement 1 is correct.
Statement 2: If a, b, and c are three different natural numbers such that a+b+c=abc then the average of a, b, and c is 3.
⇒ Let's analyze the equation a+b+c=abc
⇒ If we assume a,b,c are the smallest natural numbers that satisfy this equation, we can try a=1,b=2,c=3
⇒ 1+2+3=6 and 1×2×3=6
This satisfies the equation.
⇒ Now, calculating the average: 1 + 2 + 3 / 3 = 6/3 = 2
Conclusion: Statement 2 is incorrect because the average is 2, not 3.
Final Conclusion:
Statement 1 is correct.
Statement 2 is incorrect.
Thus, the correct answer is: 1 only
सूची - I में चार वर्ण एवं सूची - II में उनके उच्चारण-स्थान दिए गए हैं। सूची - I का मिलान, सूची - II से कीजिए तथा नीचे दिए गए कूट से सही उत्तर चुनिए :
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सूची - I |
सूची - II |
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(a) |
इ |
(i) |
कण्ठ |
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(b) |
घ |
(ii) |
ओष्ठ |
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(c) |
ब |
(iii) |
मूर्धा |
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(d) |
र |
(iv) |
तालु |
Answer (Detailed Solution Below)
CSAT Question 14 Detailed Solution
Download Solution PDFसही विकल्प 4 है।Key Points
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सूची - I |
सूची - II |
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(a) |
इ |
(i) |
तालु |
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(b) |
घ |
(ii) |
कण्ठ |
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(c) |
ब |
(iii) |
ओष्ठ |
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(d) |
र |
(iv) |
मूर्धा |
A shopkeeper sold a product at 30% loss. Had his selling price been 150 more, he would have made a profit of 10%. What was the cost price?
Answer (Detailed Solution Below)
CSAT Question 15 Detailed Solution
Download Solution PDFThe Correct answer is Option 1.
Key PointsLet the cost price (CP) of the product be x.
⇒ Selling Price (SP) at 30% loss: SP= x− 0.30x = 0.70x
⇒ Selling Price (SP) if it was ₹150 more: New SP=0.70x+150
⇒ Selling Price for a 10% profit: New SP=x+0.10x=1.10x
⇒ Setting the two expressions for New SP equal: 0.70x+150=1.10x
⇒ Rearranging the equation: 150=1.10x−0.70x
⇒ 150=0.40x
Solving for x: = 150/ 0.40 = 375
Thus, the cost price of the product is: 1) Rs. 375.