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 _____

\frac{1}{4} | |

\frac{1}{8} | |

\frac{1}{16} | |

\frac{1}{32} |

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

\frac{1}{8} | |

\frac{1}{6} | |

\frac{1}{4} | |

\frac{1}{2} |

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 ____

\frac{3}{16} | |

\frac{45}{236} | |

\frac{1}{4} | |

\frac{3}{4} |

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}

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 ____

South-East | |

North-East | |

West | |

North-West |

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

Then, the value of ( \otimes - \oplus)^2 is

0 | |

1 | |

4 | |

16 |

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 ____

10.08 AM | |

10.42 AM | |

11.00 AM | |

10.00 PM |

Question 6 Explanation:

LCM of (30 and 32) is 480

480 seconds = 8 minutes

Hence, time will be 10.08 pm

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:

2 | |

3 | |

4 | |

5 |

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

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:

The number of distinct seating arrangements possible is:

4 | |

8 | |

10 | |

16 |

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

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 .

225 | |

600 | |

1900 | |

2500 |

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

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

The new ratio of hens:ducks:goats in farm P is

5:07:13 | |

5:14:13 | |

10:21:26 | |

21:10:26 |

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}

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.

Answer to 113 question is option b plz correct it

Thank You Hania,

We have updated the answer.

for question 22 the options dosent has the right option it shall 80/81 rather than 81/80

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