CAT Quant Questions with Video Solutions

Note: These Quant questions have been selected from 1000+ CAT Quant Practice Problems with video solutions of Bodhee Prep’s Online CAT Quant Course
Question 1:
ABC is an isosceles triangle with sides AB=AC=5. D is a point in between B and C such that BD=2 and DC=4.5. Find the length of AD.
Topic: triangles

[1] \(2\sqrt 2 \)
[2] \(3\)
[3] \(4\)
[4] \(2\sqrt 3 \)

Question 2:
Let \(P=\frac{1}{{{10}^{2}}+1}+\frac{2}{{{10}^{2}}+2}+\frac{3}{{{10}^{2}}+3}+...+\frac{10}{{{10}^{2}}+10}\) then which of the following is the best approximate value of P.
Topic: series

[1] \(0.42\)
[2] \(0.52\)
[3] \(0.57\)
[4] \(0.62\)

Question 3:
How many rectangles can be formed by taking the four vertices of 18-sided regular polygon.
Topic: permutation and combination

Question 4:
Find the smallest number which has 6 distinct factors
Topic: factors

Question 5:
If Rohit drives at 20kmph, he reaches office at 3 pm. If he drives at 30kmph, he reaches office at 11 am. At what speed he should drive if he wishes to reach office at 1 pm.
Topic: speed time and distance

[1] 25 kmph
[2] 24 kmph
[3] 27 kmph
[4] None of these

CAT Quant Practice Sets [Video Explanations]


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6 thoughts on “CAT Quant Questions with Video Solutions

  1. do you have exclusive PACKAGE OF video solutions of last 10-15 years CAT EXAMS? I AM INTERESTED IN JUST THAT. I AM HELPING A GIRL APPEARING FOR CAT 2019 EXAM

    • We have already included all the good questions from CAT and other MBA entrance exams in our course.
      All these questions are with Video explanations

  2. Sir, for question no. 23:- we can do as x+y=2-z
    => cubing both sides:- x3+y3+z3=8-(2-z)(6z+3xy)
    =>as given that x3+y3+z3=8, then (2-z)(6z+3xy)=0 => z=2(considering an integer value for easy output) ,now putting z value in every eqn given :- x+y=0
    x2+y2=2
    x3+y3=0
    from the above three eqns we find that if one of x or y is +ve then another ll be -ve but both ll be of same magnitude i.e. (+-)1….thus x4+y4+z4=18

  3. Set 1 Question 5.

    I want to know the below logic would be wrong.

    Distance is constant. If the Speed increases by 10km/hr, the time decreases by 4 hours.
    So to decrease time by 2 hours, Speed can be increased by 5km/hr.

    20 + 5= 25 kmph.

    I understand something might be wrong with this logic but could someone help pinpoint that?

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