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 16:
Let \(N = 1! \times 2! \times 3! \times ..... \times 99! \times 100!\), and if \(\frac{N}{{p!}}\) is a perfect square for some positive integer \(p \le 100\), then find the value of p.
Topic: factorials

Question 17:
If \({x^2} + {y^2} = 1\) , find the maximum value of \({x^2} + 4xy - {y^2}\)
Topic: maxima minima

[1] \(1\)
[2] \(\sqrt 2 \)
[3] \(\sqrt 5 \)
[4] \(4\)

Question 18:
The compound interest on a certain amount for two years is Rs. 291.2 and the simple interest on the same amount is Rs. 280. If the rate of interest is same in both the cases, find the Principal amount
Topic: sici

[1] 1200
[2] 1400
[3] 1700
[4] 1750

Question 19:
In the diagram given below, the circle and the square have the same center O and equal areas. The circle has radius 1 and intersects one side of the square at P and Q. What is the length of PQ?

Topic: circles

[1] 1
[2] 3/2
[3] \(\sqrt {4 - \pi } \)
[4] \(\sqrt {\pi  - 1} \)

Question 20:
What is the remainder when \({{x}^{276}}+12\) is divided by \({{x}^{2}}+x+1\) given that the remainder is a positive integer?
Topic: remainders

CAT Quant Practice Sets [Video Explanations]


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CAT Quant Questions Set 05
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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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