Numerical Ability

Question 1
Consider a square sheet of side 1 unit. The sheet is first folded along the main diagonal. This is followed by a fold along its line of symmetry. The resulting folded shape is again folded along its line of symmetry. The area of each face of the final folded shape, in square units, equal to _____
A
\frac{1}{4}
B
\frac{1}{8}
C
\frac{1}{16}
D
\frac{1}{32}
GATE ME 2021 SET-2   General Aptitude
Question 1 Explanation: 


Area=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}
Question 2


The ratio of the area of the inscribed circle to the area of the circumscribed circle of an equilateral triangle is
A
\frac{1}{8}
B
\frac{1}{6}
C
\frac{1}{4}
D
\frac{1}{2}
GATE ME 2021 SET-2   General Aptitude
Question 2 Explanation: 


\begin{aligned} \sin 30^{\circ} &=\frac{r}{R} \\ \text { Area ratio } &=\frac{\pi r^{2}}{\pi R^{2}}=\sin ^{2} 30=\frac{1}{4} \end{aligned}
Question 3
A box contains 15 blue balls and 45 black balls. If 2 balls are selected randomly, without replacement, the probability of an outcome in which the first selected is a blue ball and the second selected is a black ball, is ____
A
\frac{3}{16}
B
\frac{45}{236}
C
\frac{1}{4}
D
\frac{3}{4}
GATE ME 2021 SET-2   General Aptitude
Question 3 Explanation: 
The probability of first ball is blue and second ball is black is given as,
P=\frac{15}{60} \times \frac{45}{59}=\frac{45}{236}
Question 4
The front door of Mr. X's house faces East. Mr. X leaves the house, walking 50 m straight from the back door that is situated directly opposite to the front door. He then turns to his right, walks for another 50 m and stops. The direction of the point Mr. X is now located at with respect to the starting point is ____
A
South-East
B
North-East
C
West
D
North-West
GATE ME 2021 SET-2   General Aptitude
Question 4 Explanation: 


Question 5
If \oplus \div \odot =2;\;\oplus \div \triangle =3;\;\odot +\triangle =5;\;\triangle \times \otimes =10,
Then, the value of ( \otimes - \oplus)^2 is
A
0
B
1
C
4
D
16
GATE ME 2021 SET-2   General Aptitude
Question 5 Explanation: 
\begin{aligned} \frac{\oplus}{\odot}&=2, \frac{\oplus}{\Delta}=3 \\ \therefore \qquad\frac{\Delta}{\odot}&=\frac{2}{3} &\ldots(1)\\ \odot+\Delta&=5 &\ldots(2)\\ \text{From (1) and (2)}\\ \Delta&=2, \odot=3\\ \text{and}\\ \oplus &=6,2 \times \otimes=10 \\ \otimes &=5 \\ \Rightarrow \quad(\otimes-\oplus)^{2}&=(5-6)^{2} =1 \end{aligned}
Question 6
A digital watch X beeps every 30 seconds while watch Y beeps every 32 seconds. They beeped together at 10 AM. The immediate next time that they will beep together is ____
A
10.08 AM
B
10.42 AM
C
11.00 AM
D
10.00 PM
GATE ME 2021 SET-2   General Aptitude
Question 6 Explanation: 
LCM of (30 and 32) is 480
480 seconds = 8 minutes
Hence, time will be 10.08 pm
Question 7
Five persons P, Q, R, S and T are to be seated in a row, all facing the same direction, but not necessarily in the same order. P and T cannot be seated at either end of the row. P should not be seated adjacent to S. R is to be seated at the second position from the left end of the row. The number of distinct seating arrangements possible is:
A
2
B
3
C
4
D
5
GATE ME 2021 SET-2   General Aptitude
Question 7 Explanation: 
The possible distinct arrangement are
S R P T A, A R P T S, S R T P A
Hence, number of distinct sitting arrangement. = 3
Question 8
Five persons P, Q, R, S and T are sitting in a row not necessarily in the same order. Q and R are separated by one person, and S should not be seated adjacent to Q.
The number of distinct seating arrangements possible is:
A
4
B
8
C
10
D
16
GATE ME 2021 SET-1   General Aptitude
Question 8 Explanation: 
The possible seating arrangements are
QPRST, QTRSP
QPRTS, QTRPS
PQTRS, TQPRS
SPQTR, STQPR
RPQTS, RTQPS
SRPQT, SRTQP
SPRTQ, STRPQ
PSRTQ, TSRPQ
Hence, total seating arrangements are 16
Question 9


The distribution of employees at the rank of executives, across different companies C1, C2,... ,C6 is presented in the chart given above. The ratio of executives with a management degree to those without a management degree in each of these companies is provided in the table above. The total number of executives across all companies is 10,000.

The total number of management degree holders among the executives in companies C2 and C5 together is .
A
225
B
600
C
1900
D
2500
GATE ME 2021 SET-1   General Aptitude
Question 9 Explanation: 
Number of employee in C2 company =\frac{5}{100} \times 10000=500
Number of management degree holder employee in \mathrm{C} 2=\frac{1}{5} \times 500=100
Number of employee in \mathrm{C} 5 company =\frac{20}{100} \times 10000=2000
Number of management degree holder employee in \mathrm{C} 5=\frac{9}{10} \times 2000=1800
Total management degree holder employee =100+1800=1900
Question 10
The number of hens, ducks and goats in farm P are 65, 91 and 169, respectively. The total number of hens, ducks and goats in a nearby farm Q is 416. The ratio of hens:ducks:goats in farm Q is 5:14:13. All the hens, ducks and goats are sent from farm Q to farm P.
The new ratio of hens:ducks:goats in farm P is
A
5:07:13
B
5:14:13
C
10:21:26
D
21:10:26
GATE ME 2021 SET-1   General Aptitude
Question 10 Explanation: 
In farm P,
Hens =65, Ducks =91, Goats =169
In farm Q,
Hens : Ducks : Goats
5: 14: 13
\begin{aligned} \text { Hens } &=\frac{5}{32} \times 416=65 \\ \text { Ducks }&=\frac{14}{32} \times 416=182\\ \text { Goats }&=\frac{13}{32} \times 416=169 \end{aligned}
\because From farm d, hens, ducks and goats are sent to farm P.
\begin{aligned} \therefore \text{ Total hens }&=65+65=130\\ \text { Total ducks } &=91+182=273 \\ \text { Total goats } &=169+169=338 \\ \text { New ratio } &=130: 273: 338 \\ &=10: 21: 26 \end{aligned}


There are 10 questions to complete.

4 thoughts on “Numerical Ability”

  3. mistake in question 25 , as per solution the data for april second row (second row last column) should be 70 but here its given as 170

    Reply

Leave a Comment