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?

  1. There is only one such number.
  2. There are only two such numbers.
  3. There can be more than one such number.
  4. No such number exists.

Answer (Detailed Solution Below)

Option 3 : There can be more than one such number.

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:

  1. I only
  2. II only
  3. Both I and II
  4. Neither I nor II

Answer (Detailed Solution Below)

Option 1 : I only

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?

  1. k
  2. k - 1
  3. k + 1
  4. k/2

Answer (Detailed Solution Below)

Option 1 : k

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?

  1. Value-I < Value-II
  2. Value-II < Value-I
  3. Value-I = Value-II
  4. Cannot be determined due to insufficient data

Answer (Detailed Solution Below)

Option 3 : Value-I = Value-II

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?

  1. Value-I < Value-II
  2. Value-II < Value-I
  3. Value-I = Value-II
  4. Cannot be determined due to insufficient data

Answer (Detailed Solution Below)

Option 3 : Value-I = Value-II

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 :

सूची - I

सूची - II

(a)

कक्ष

(i)

व्यक्तिवाचक संज्ञा

(b)

कक्षा

(ii)

भाववाचक संज्ञा

(c)

सजावट

(iii)

जातिवाचक संज्ञा

(d)

महाभारत

(iv)

समूहवाचक संज्ञा

  1. (a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)
  2. (a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)
  3. (a) - (iv), (b) - (iii), (c) - (i), (d) - (ii)
  4. (a) - (i), (b) - (iv), (c) - (ii), (d) - (iii)

Answer (Detailed Solution Below)

Option 2 : (a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)

CSAT Question 6 Detailed Solution

Download Solution PDF

सही विकल्प 2 है।Key Pointsसही मिलान होगा-

सूची - I

सूची - II

(a)

कक्ष

(i)

जातिवाचक संज्ञा

(b)

कक्षा

(ii)

समूहवाचक संज्ञा 

(c)

सजावट

(iii)

भाववाचक संज्ञा

(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 :

सूची - I

सूची - II

(a)

लता

(i)

तद्भव शब्द

(b)

पत्ता

(ii)

तत्सम शब्द

(c)

जूता

(iii)

विदेशज शब्द

(d)

गुलाब

(iv)

देशज शब्द

  1. (a) - (iv), (b) - (ii), (c) - (iii), (d) - (i)
  2. (a) - (iii), (b) - (ii), (c) - (i), (d) - (iv)
  3. (a) - (ii), (b) - (i), (c) - (iv), (d) - (iii)
  4. (a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)

Answer (Detailed Solution Below)

Option 3 : (a) - (ii), (b) - (i), (c) - (iv), (d) - (iii)

CSAT Question 7 Detailed Solution

Download Solution PDF

सही विकल्प 3 है।Key Pointsसही मिलान है-

सूची - I

सूची - II

(a)

लता

(i)

तत्सम शब्द 

(b)

पत्ता

(ii)

तद्भव शब्द

(c)

जूता

(iii)

देशज शब्द

(d)

गुलाब

(iv)

विदेशज शब्द

The remainder, when 1 + (1 × 2) + (1 × 2 × 3) +......+ (1 × 2 × 3 × .... × 500) is divided by 8, is 

  1. 1
  2. 2
  3. 3
  4. 4

Answer (Detailed Solution Below)

Option 1 : 1

CSAT Question 8 Detailed Solution

Download Solution PDF

The 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.

A nail is nailed into the trunk of a tree at a point 1 m from the soil level. After 2 years the nail will

  1. Remain at same position
  2. Move sideways
  3. Move up
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : Remain at same position

CSAT Question 9 Detailed Solution

Download Solution PDF

The Correct answer is Option 1. 

Key Points

  • A tree’s trunk grows primarily from the tips (apical meristems) rather than pushing older wood upward. Although the trunk thickens (increases in girth) over time due to lateral growth, any fixed point along its length does not shift significantly up or down.
  • Therefore, a nail driven into the trunk at 1 m above the ground will still be at roughly the same height two years later, even though it may become more embedded as the trunk thickens. Hence Correct answer is Option 1. 

At what time between 2 and 3 O'Clock will the hands of a clock be together?

  1. \(38\frac{2}{11}\) minutes past 2
  2. \(38\frac{10}{11}\) minutes past 2 
  3. \(10\frac{10}{43}\) minutes past 2
  4. \(10\frac{10}{11}\) minutes past 2

Answer (Detailed Solution Below)

Option 4 : \(10\frac{10}{11}\) minutes past 2

CSAT Question 10 Detailed Solution

Download Solution PDF

The Correct answer is Option 4. 

Key Points

Express each hand’s angle from 12 o’clock:

Minute hand speed: 6° per minute.

Hour hand speed: 0.5° per minute, with a 60° head start at 2:00.

Set up the equation for coincidence:

Let t be the number of minutes after 2:00. We want the minute hand’s angle to equal the hour hand’s angle:

6t  =  60  +  0.5t

Simplifying:

6t − 0.5t = 60 → 5.5t = 60 → t = 60 / 5.5 = 120 / 11 = 10 10⁄11 minutes

Therefore, the hands coincide at 10 10⁄11 minutes past 2, 

Hence Correct answer is Option 4. 

- amglogisticsinc.net

 

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?

  1. 35
  2. 40
  3. 49
  4. 55

Answer (Detailed Solution Below)

Option 3 : 49

CSAT Question 11 Detailed Solution

Download Solution PDF

The 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 ?

  1. 40.5
  2. 42.5
  3. 44.5
  4. 46.5

Answer (Detailed Solution Below)

Option 4 : 46.5

CSAT Question 12 Detailed Solution

Download Solution PDF

The 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

 

How many triangles are there in the following figure?

qImage6792009a161e26e2a339a779

  1. 12
  2. 14
  3. 10
  4. 8

Answer (Detailed Solution Below)

Option 1 : 12

CSAT Question 13 Detailed Solution

Download Solution PDF

The Correct answer is Option 1. 

Key Points

Identify the small “inner” triangles:

There are 6 small triangles formed directly by the intersecting lines.

Combine small triangles to form larger ones:

You can form 3 “medium” triangles by joining pairs of those small triangles.

You can form 2 more “medium” triangles by joining three small triangles at once.

Include the outermost triangle: The large boundary triangle itself counts as 1 more.

When you add them all up:

6 (small) + 3 (medium, using 2 small ) + 2 (medium, using 3 small ) + 1 (largest outer triangle) = 12 Triangles 

Hence Correct answer is Option 1. 

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 :

सूची - I

सूची - II

(a)

(i)

कण्ठ

(b)

(ii)

ओष्ठ

(c)

(iii)

मूर्धा

(d)

(iv)

तालु

  1. (a) - (i), (b) - (iv), (c) - (iii), (d) - (ii)
  2. (a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)
  3. (a) - (iii), (b) - (ii), (c) - (i), (d) - (iv)
  4. (a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)

Answer (Detailed Solution Below)

Option 4 : (a) - (iv), (b) - (i), (c) - (ii), (d) - (iii)

CSAT Question 14 Detailed Solution

Download Solution PDF

सही विकल्प 4 है।Key Points

सूची - I

सूची - II

(a)

(i)

तालु

(b)

(ii)

कण्ठ 

(c)

(iii)

ओष्ठ 

(d)

(iv)

मूर्धा

Raj travelled from a point X straight to Y at a distance of 80 metres. He turned right and walked 50 metres, then again turned right and walked 70 metres. Finally, he turned right and walked 50 metres. How far is he from the starting point?

  1. 30 metres
  2. 45 metres
  3. 20 metres
  4. 10 metres

Answer (Detailed Solution Below)

Option 4 : 10 metres

CSAT Question 15 Detailed Solution

Download Solution PDF

The Correct answer is Option 4. 

Key Points

Here you can Understand through Given Diagram 

Screenshot 2025-02-10 124004

Get Free Access Now
Hot Links: real teen patti teen patti joy official yono teen patti