# Numerical Ability

 Question 1
In an equilateral triangle PQR, side PQ is divided into four equal parts, side QR is divided into six equal parts and side PR is divided into eight equals parts. The length of each subdivided part in cm is an integer. The minimum area of the triangle PQR possible, in $cm^2$, is
 A 18 B 24 C $48 \sqrt{3}$ D $144 \sqrt{3}$
GATE CE 2021 SET-2   General Aptitude
Question 1 Explanation: For $\left(\frac{a}{4}, \frac{a}{6}, \frac{a}{8}\right)$ to be integer, a must be LCM of 4, 6 and 8. So a = 24
$\text { Area }=\frac{\sqrt{3}}{4} a^{2}=\frac{\sqrt{3}}{4} \times 24^{2}=144 \sqrt{3}$
 Question 2 In the figure shown above, PQRS is a square. The shaded portion is formed by the intersection of sectors of circles with radius equal to the side of the square and centers at S and Q.
The probability that any point picked randomly within the square falls in the shaded area is __________
 A $4-\frac{\pi}{2}$ B $\frac{1}{2}$ C $\frac{\pi}{2}-1$ D $\frac{\pi}{4}$
GATE CE 2021 SET-2   General Aptitude
Question 2 Explanation:
\begin{aligned} \text { Probability } &=\frac{f A}{T A} \\ f A &=\left(\frac{\pi r^{2}}{4}-\frac{r^{2}}{2}\right) \times 2 \\ \frac{f A}{T A} &=\frac{\left(\frac{\pi r^{2}}{4}-\frac{r^{2}}{2}\right) \times 2}{r^{2}}=\left(\frac{\pi}{2}-1\right) \end{aligned}
 Question 3
On a planar field, you travelled 3 units East from a point O. Next you travelled 4 units South to arrive at point P. Then you travelled from P in the North-East direction such that you arrive at a point that is 6 units East of point O. Next, you travelled in the North-West direction, so that you arrive at point Q that is 8 units North of point P.
The distance of point Q to point O, in the same units, should be _____________
 A 3 B 4 C 5 D 6
GATE CE 2021 SET-2   General Aptitude
Question 3 Explanation: $O Q=\sqrt{3^{2}+4^{2}}=5$
 Question 4
Four persons P, Q, R and S are to be seated in a row. R should not be seated at the second position from the left end of the row. The number of distinct seating arrangements possible is:
 A 6 B 9 C 18 D 24
GATE CE 2021 SET-2   General Aptitude
Question 4 Explanation:
Number of arrangements $=3 \times 3 !=18$
 Question 5
$\oplus$ and $\odot$ are two operators on numbers p and q such that $p \odot q=p-q$, and $p \oplus q=p \times q$
Then, $(9 \odot(6 \oplus 7)) \odot(7 \oplus(6 \odot 5))=$
 A 40 B -26 C -33 D -40
GATE CE 2021 SET-2   General Aptitude
Question 5 Explanation:
\begin{aligned} [9-(6 \times 7)]-[7 \times 1] &=-33-7 \\ &=-40 \end{aligned}
 Question 6
Two identical cube shaped dice each with faces numbered 1 to 6 are rolled simultaneously. The probability that an even number is rolled out on each dice is:
 A $\frac{1}{36}$ B $\frac{1}{12}$ C $\frac{1}{8}$ D $\frac{1}{4}$
GATE CE 2021 SET-2   General Aptitude
Question 6 Explanation:
Probability of getting even number on a dice$=\frac{3}{6}=\frac{1}{2}$
$\therefore$Two dice are rolled simultaneously,
Hence required probability $=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$
 Question 7 The mirror image of the above text about X-axis is A A B B C C D D
GATE CE 2021 SET-2   General Aptitude
 Question 8
A function, $\lambda$, is defined by
$\lambda(p, q)=\left\{\begin{array}{cl} (p-q)^{2}, & \text { if } p \geq q \\ p+q, & \text { if } p \lt q \end{array}\right.$
The value of the expression $\frac{\lambda(-(-3+2),(-2+3))}{(-(-2+1))}$ is:
 A -1 B 0 C $\frac{16}{3}$ D 16
GATE CE 2021 SET-1   General Aptitude
Question 8 Explanation:
$\frac{\lambda(-(-3+2),(-2+3))}{(-(2+1))}=\lambda \frac{(1,1)}{1}=\lambda(1,1)$
So, 1st definition will be applicable as p = q.
$\text { Hence, } \qquad \lambda(1,1)=(1-1)^{2}=0$
 Question 9 Five line segments of equal lengths, PR, PS, QS, QT and RT are used to form a star as shown in the figure above.
The value of $\theta$, in degrees, is ________
 A 36 B 45 C 72 D 108
GATE CE 2021 SET-1   General Aptitude
Question 9 Explanation: Sum of angle formed at the pentagon = $540^{\circ}$
Each angle of $=\frac{540}{5}=108^{\circ}$
$\angle x=180-108=72^{\circ}$
Sum of angle of triangle $=180^{\circ}$
\begin{aligned} 72^{\circ}+72^{\circ}+\theta &=180^{\circ} \\ \theta &=36^{\circ} \end{aligned}
 Question 10
Consider two rectangular sheets, Sheet M and Sheet N of dimensions 6cm x 4cm each.
Folding operation 1: The sheet is folded into half by joining the short edges of the current shape.
Folding operation 2: The sheet is folded into half by joining the long edges of the current shape.
Folding operation 1 is carried out on Sheet M three times.
Folding operation 2 is carried out on Sheet N three times.
The ratio of perimeters of the final folded shape of Sheet N to the final folded shape of Sheet M is ____.
 A 0.546528 B 0.126389 C 0.295139 D 0.217361
GATE CE 2021 SET-1   General Aptitude
Question 10 Explanation: $(\text { Perimeter })_{M}=2(2+1.5)=7$ $\text { (Perimeter })_{N}=2(0.5+6)=13$
Required ratio $=\frac{13}{7}$

There are 10 questions to complete.

### 1 thought on “Numerical Ability”

1. And solution?? 