CMAT 2026 Previous Year Paper

Let us assume that the number of students in Kendriya Vidyalaya = 100.
This means that the number of students in Navodaya Vidyalaya = 1.6*100 = 160.
Athletes in Kendriya Vidyalaya = 20% of 100 = 20 and,
Athletes in Navodaya Vidyalaya = 25% of 160 = 40.
Required percentage = $$\frac{40}{20}\times\ 100\ =\ 200\%$$

Only statements A and C are correct i.e. Option D.

Let us assume the height and radius to be R.
Volume of the cylinder = $$\pi\ \times\ R^2\times\ height\ =\ \pi\ \times\ R^3$$
==> $$\frac{22}{7}\times\ R^3=\frac{176}{7}$$
This gives R = 2 cm i.e. Option C.

Triangle CDA will be similar to Triangle CAB with AAA similarity.
This gives : $$\frac{CD}{CA}=\frac{CA}{CB}$$
==> $$\frac{2}{4}=\frac{4}{BC}$$
==>BC = 8 units i.e. Option A.

Let us assume the cost of 2 watches to be 100a and 100b.
100a + 100b = 1000 ==> a + b = 10.
Selling price of 1st watch after 5% loss = 0.95*100a = 95a.
Selling price of 2nd watch after 20% gain = 1.20*100b = 120b.
Total selling price => 95a + 120b = 1050.
We will multiple the 1st equation with 95 and subtract it from the 2nd:
25b = 100 ==> b = 4 and a = 6.
The Costs of the 2 watches come out to be 600 and 400 i.e. Option C.

Quantity of milk in the mixture = $$\frac{27}{27+7}\times\ 85\ =\ 67.5$$.
Let us assume that X litres of water is added which makes the quantity of new solution = 85+X.
Milk content remains the same which means that:
$$\frac{3}{3+1}\times\ \times\ \left(85+X\right)=67.5$$.
This gives X = 5 litres.

The nth term of the series can be written as:
$$\frac{2n+1}{n^2\left(n+1\right)^2}$$.
Now, 2n+1 can be written as: $$\left(n+1\right)^2-n^2$$.
This gives nth term to be:
$$\frac{\left(\left(n+1\right)^2-n^2\right)}{n^2\times\ \left(n+1\right)^2}=\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}$$.
Now, we put values of n from 2 to 7 and add them to get:
$$\left(\frac{1}{2^2}-\frac{1}{3^2}\right)+\left(\frac{1}{3^2}-\frac{1}{4^2}\right)+...\left(\frac{1}{7^2}-\frac{1}{8^2}\right)=\frac{1}{2^2}-\frac{1}{8^2}$$.
==> $$\frac{1}{4}-\frac{1}{64}=\frac{15}{64}$$.

Let the odd numbers be: 2n+1, 2n+3, 2n+5.
Their sum will be: 2n+1 + 2n+3 + 2n+5 = 6n+9
==> 3*(2n+3).
So, the sum is always divisible by 3 only i.e. Option D.

Probability of 2 consecutive heads being never together = 1 - Probability of 2 consecutive heads being together.
Cases where 2 heads are consecutive :
HHHH = 1 case.
THHH, HTHH, HHTH and HHHT = 4 cases.
HHTT, THHT and TTHH = 3 cases.
Total cases where 2 heads come consecutively = 8 cases.
Total number of possible cases with 4 coins = 2*2*2*2 = 16.
Probability of 2 consecutive heads = 8/16 = 1/2.
Now, Probability of 2 consecutive heads being never together = 1 - 1/2 = 1/2 i.e. Option D.

Let us assume $$\sqrt{\ \sqrt{\ 5}+2}+\sqrt{\ \sqrt{\ 5}-2}=a$$.
Squaring both sides:
$$\left(\sqrt{\ 5}+2\right)+\left(\sqrt{\ 5}-2\right)+2\sqrt{\ \left(\sqrt{\ 5}+2\right)\left(\ \sqrt{\ 5}-2\right)}=2\sqrt{\ 5}+2=a^2$$
==> $$a\ =\ \sqrt{\ 2}\times\ \left(\sqrt{\ \sqrt{\ 5}+1}\right)$$
and, $$\frac{a}{\left(\sqrt{\ \sqrt{\ 5}+1}\right)}=\sqrt{\ 2}$$ i.e. Option B.