Staff Selection Commission Combined Higher Secondary Level Exam
Question : Two sides of a parallelogram are 20 cm and 25 cm. If the altitude corresponding to the side of length 25 cm is 10 cm, then the altitude corresponding to the other pair of sides is:
Option 1: 10.5 cm
Option 2: 12 cm
Option 3: 12.5 cm
Option 4: 10 cm
Correct Answer: 12.5 cm
Solution : Given: Two sides of a parallelogram are 20 cm and 25 cm. If the altitude corresponding to the side of length 25 cm is 10 cm. Let the altitude corresponding to the other pair of sides be $h$ cm. According to the question, 20
Question : AB and AC are tangents to a circle with centre O. A is the external point of the circle. The line AO intersects the chord BC at D. The measure of the $\angle$BDO is:
Option 1: 60°
Option 2: 90°
Option 3: 45°
Option 4: 75°
Correct Answer: 90°
Solution : AO intersects the chord BC at D. In $\triangle$ ABD and $\triangle$ ACD AB = AC = tangents from an exterior point A $\angle$ BAD = $\angle$CAD AD is common. So, $\angle$ BAD $\cong$ $\angle$CAD $\angle$ BDA and $\angle$ ADC are linear pair and equal.
Question : The value of $2 \frac{3}{5} \div\left[2 \frac{1}{3} \div\left\{4 \frac{1}{3}-\left(2 \frac{1}{2}+\frac{2}{3}\right)\right\}\right]$ is equal to:
Option 1: $1 \frac{3}{10}$
Option 2: $2 \frac{7}{10}$
Option 3: $2 \frac{3}{7}$
Option 4: $1 \frac{3}{7}$
Correct Answer: $1 \frac{3}{10}$
Solution : $2 \frac{3}{5} \div[2 \frac{1}{3} \div\{4 \frac{1}{3}-(2 \frac{1}{2}+\frac{2}{3})\}]$ $= 2 \frac{3}{5} \div[2 \frac{1}{3} \div\{4 \frac{1}{3}-(\frac{5}{2}+\frac{2}{3})\}]$ $= 2 \frac{3}{5} \div[2 \frac{1}{3} \div\{4 \frac{1}{3}-(\frac{15+4}{6})\}]$ $= 2 \frac{3}{5} \div[2 \frac{1}{3} \div\{\frac{13}{3}-\frac{19}{6}\}]$ $= 2 \frac{3}{5} \div[2 \frac{1}{3} \div\{\frac{26-19}{6}\}]$ $= 2 \frac{3}{5} \div[\frac{7}{3} \div\frac{7}{6}]$ $= 2 \frac{3}{5} \div[\frac{7}{3} \times\frac{6}{7}]$ $=
Question : Directions: If DEAR is coded as 6 – 8 – 3 – 21, how will you code TRACK?
Option 1: 22 – 21 – 3 – 6 – 11
Option 2: 22 – 21 – 3 – 6 – 13
Option 3: 22 – 21 – 4 – 5 – 10
Option 4: 20 – 21 – 3 – 6 – 17
Correct Answer: 22 – 21 – 3 – 6 – 13
Solution : Given: DEAR is coded as 6 – 8 – 3 – 21
Add 2 and 3 alternatively to the place values of the letters of DEAR to get the required code – D(4) + 2 = 6;
Question :
Five Kingdom classifications were given by:
Option 1:
Whittaker
Option 2:
Haeckel
Option 3: Linnaeus
Option 4: Copeland
Correct Answer:
Solution : The correct option is Whittaker.
In 1969, R. H. Whittaker proposed the five-kingdom division. The Monera, Protista, Fungi, Plantae and Animalia are the five kingdoms that make up the Five Kingdom Classification system. The organisms included in this classification are divided based on evolutionary
Question : Name the Prime Minister who brought about a thaw in India-China relations by signing the "Line of Actual Control".
Option 1: Lal Bahadur Shastri
Option 2: P.V. Narashima Rao
Option 3: Chandrasekhar
Option 4: V.P. Singh
Correct Answer: P.V. Narashima Rao
Solution : The correct option is P.V. Narashima Rao
The signing of the "Line of Actual Control" by Prime Minister Narasimha Rao resulted in the improvement of ties between India and China. He served as India's Prime Minister from 1991 to 1996. During Rao's tenure
Question : Solve (0.125 × 0.01) ÷ (0.5 × 0.005).
Option 1: 0.6
Option 2: 0.5
Option 3: 0.8
Option 4: 0.7
Correct Answer: 0.5
Solution : $(0.125× 0.01)÷ (0.5 × 0.005)$ $=0.00125\ ÷\ 0.0025$ $=0.5$ Hence, the correct answer is 0.5.
Question : In a circle if PQ is the diameter of the circle and R is on the circumference of the circle such that $\angle$PQR = 30°, then $\angle$RPQ =?
Option 1: 90°
Option 2: 60°
Option 3: 30°
Option 4: 45°
Correct Answer: 60°
Solution : Given: $\angle$PQR = 30° We know the angle of a semi-circle is a right angle, so $\angle$PRQ = 90° So, $\angle$RPQ = 180° – $\angle$PQR – $\angle$PRQ = 180° – 30° – 90° = 60° Hence, the correct answer is 60°.
Question : Directions: If 14 March 2007 is Wednesday, then what will be the day of the week on 12 April 2013?
Option 1: Tuesday
Option 2: Friday
Option 3: Monday
Option 4: Sunday
Correct Answer: Friday
Solution : Given: 14 March 2007 is a Wednesday.
Let's calculate the total number of days from the 14th of March 2007 to the 12th of April 2013 –
Number of days from 14 March 2007 to 31 December 2007 = 17 + 30 + 31 +
Question : Directions: The following table gives the population of a locality from 1988 to 1992. Read the table and answer the question.
Increase(+)decrease(-)
over preceding year
The number of children in 1989 was:
Option 1: 25,670
Option 2: 14,040
Option 3: 13,970
Option 4: 15,702
Correct Answer: 25,670
Solution : Total population in 1989 = 1,46,947 + 11,630 = 1,58,577 The number of children in 1989 = 1,58,577 – 70,391 – 62,516 = 25,670 Hence, the correct answer is 25,670.
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