CAT Quant Questions with Video Solutions

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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 6:
A five digit number N has all digits different and contains digits 1,3,4,5, and 6 only. If N is the smallest possible number such that it is divisible by 11, then what is the tens place digit of N.
Topic: divisibility

[1] 1
[2] 3
[3] 4
[4] 5

Question 7:
If \(\alpha \) and \( - \alpha \) are the roots of the equation \(2{x^3} - 5{x^2} - 8x + n = 0\) . Find the value of n?
Topic: equations

Question 8:
The bases of a trapezoid have lengths 10 and 15. A segment parallel to the bases passes through the point of intersection of the diagonals and extends from one side to the other. Find the length of the segment
Topic: quadrilaterals

[1] 5
[2] \(\sqrt {150} \)
[3] \(\frac{{25}}{{\sqrt 2 }}\)
[4] 12

Question 9:
The value of a fraction represented in base n is \(0.\overline {111} \) while in base 2n it takes the simpler form 0.2n. What is n?
Topic: base system

Question 10:
If Gopal and Harish can complete a work together in 20 days, and Gopal alone can do the same work in 32 days. When both work together their efficiencies reduce by 20% compare to the efficiency when they would have worked alone. Find the number of days Harish alone takes to complete the same work.
Topic: work and time

CAT Quant Practice Sets [Video Explanations]


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8 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

  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?

    • the assumption of “to decrease time by 2 hours, speed can be increase by 5km/hr” is wrong. it would be right if you know the initial time and consider from the start but since the journey is already going on, u can’t do that.

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