CMAT 2026 Previous Year Paper

Number of cars sold by P in February, April and May = 94 + 108 + 98 = 300.
Number of cars sold by S in March, May and June = 90 + 89 + 91 = 270.
The required ratio = 300 : 270 = 10:9 i.e. Option D. 

Total number of cars sold by Dealer P from Feb to June = 94 + 85 + 108 + 98 + 95 = 480.
Total cars sold by all the dealers in June = 396.
The required percentage = $$\frac{480-396}{396}\times\ 100\ \approx\ \ 21.2\%$$.

The total number of cars sold by dealer Q in April, May and June = 97 + 102 + 108 = 307.
the total number of cars sold by all dealers in February and April = 400 + 400 = 800.
The required percentage = $$\frac{307}{800}\times\ 100\ =\ 38\frac{3}{8}\%$$ i.e. Option D.

The unit digit of the expression will be equal to the unit digit of:
$$48\times\ 4^{100}+4^{101}$$.
Odd powers of 4 gives 4 as the unit digit and even powers give 6.
So, the unit digit will be the unit digit of 8*6 + 4 i.e. 2.

$$B+\frac{1}{C-9}=\frac{B\left(C-9\right)+1}{C-9}$$
$$\frac{1}{B+\frac{1}{C-9}}=\frac{C-9}{B\left(C-9\right)+1}$$.
Putting this value in the equation to get:
$$A+\frac{C-9}{B\left(C-9\right)+1}=\frac{A\cdot B\cdot\left(C-9\right)+A+\left(C-9\right)}{B\cdot\left(C-9\right)+1}=\frac{29}{5}$$.
This means that B(C-9) + 1 = 5 ==> B(C-9) = 4.
C- 9 can be 1 or 2 or 4.
and, $$A\cdot B\cdot\left(C-9\right)+A+\left(C-9\right)=29$$.
4*A + A + (C-9) = 29
5A + (C-9) = 29
C-9 has to be 4 which gives A = 5 and C = 13.
B*(C-9) = 4 which means B = 1.
A + B + C = 5 + 1 + 13 = 19.

Area of (I) = Length * Breadth = 9*4 = 36 sq cm.
Perimeter of (I) = 2*(Length + Breadth) = 2*(9+4) = 26 cm.

Area of (II) = Square of side = 6^2 = 36 sq cm.
Perimeter of (II) = 4*side = 4*6 = 24 cm.

Area of (III) = half of Perpendicular * Base = half of 9*8 = 36 sq cm.
Perimeter of (III) = sum of sides = 9 + 8 + $$\sqrt{\ 9^2+8^2}=17\ +\ \sqrt{\ 145}$$.

Only statement B is correct i.e. The areas of three figures are equal.

Cost Price = 320 and he earned a profit of 15% which makes the Selling Price = 1.15*320 ==> 368.
This Selling Price is attained by giving a discount of 32 on the Marked Price which gives:
Marked Price = 368 + 32 = 400.
Profit earned after he sold it at Marked Price = $$\frac{400-320}{320\ }\times\ 100\ =\ 25\%$$.

Let us assume that the work to be done is 32 units i.e. the LCM of 8, 16 and 32.
Efficiencies of Raman, Satish and Ashok are 2, 4 and 1 units per day respectively.
Let us assume that they completed the work in X days.
Satish worked for 2 days, Raman worked for X-3 days and Ashok worked for X days which led to the completing of work in X days.
Total work done: 2*4 + (X-3)*2 + X*1 = 3X + 2.
This is equal to Total work = 32.
3X + 2 = 32
==> X = 10 days.

All the devices will again beep together after every N minutes where N is the LCM of 30 mis, 1hr, 1hr 30 mins and 1 hr 45 mins.
We need to find the LCM of 30, 60, 90 and 105 mins.
Their LCM will be 1260 mins i.e. 21 hrs.
So, if they beeped together at 12 noon, next time, they will beep together at 12 noon + 21 hrs = 9 am.

We will use the successive discount formula to calculate the overall discount.
Overall discount after giving a discount of 20% and 5% = $$20+5-\frac{20\times\ 5}{100}=25-1\ =24\%$$.
Now, we give another discount of 12.5% to get an overall discount of:
$$24\ +\ 12.5-\frac{24\times\ 12.5}{100}=33.5\%$$ i.e. Option A.