Staff Selection Commission Combined Higher Secondary Level Exam
Question : The successive discounts of 15% and 20% are equivalent to a single discount of:
Option 1: 25%
Option 2: 28%
Option 3: 32%
Option 4: 35%
Correct Answer: 32%
Solution : Single equivalent discount $=(a+b-\frac{a×b}{100})$%, where $a=15\%$ and $b=20$% are successive discounts. $\therefore$ Single equivalent discount $=15+20-\frac{15×20}{100}=35-\frac{300}{100}=32$% Hence, the correct answer is 32%.
Question : Comprehension:
In the following passage, some of the words have been deleted. Read the passage carefully and select the correct answer for the given blank out of the four alternatives. Each year, the U.S. Department of Agriculture and the Environmental Protection Agency (1)_____ federal pollinator experts to share the latest scientific findings on bee and pollinator health, and assess the status of (2)______ important insects, birds, bats, and other species. One clear (3)_______ from this year's meeting was that climate change (4)______ a new and formidable stressor for bees, potentially amplifying previously known issues in ways that scientists can't yet (5)______but need to prepare for.
Question Select the most appropriate option to fill in the blank number (1).
Option 1: accost
Option 2: host
Option 3: roast
Option 4: mandates
Correct Answer: host
Solution : The most appropriate option is the second option.
Explanation:
In the context of events or meetings, "host" means to organise or hold. The U.S. Department of Agriculture and the Environmental Protection Agency host federal pollinator experts, indicating that they organise and hold the annual meeting.
Question : If $xy=48$ and $x^2+y^2=100$, then $(x+y)$ is:
Option 1: 12
Option 2: 16
Option 3: 18
Option 4: 14
Correct Answer: 14
Solution : Given: $xy = 48$ and $x^{2}+y^{2} = 100$ We know the algebraic identity, $(x+y)^{2} = x^{2}+y^{2}+2xy$ By substituting the values in the above equation, we get: $⇒(x+y)^{2} = 100+2×48$ $⇒(x+y) = \sqrt{196}$ $\therefore (x+y) = 14$ Hence, the correct answer is 14.
Question : The following sentence has been divided into parts. One of them may contain an error. Select the part that contains the error from the given options. If you don't find any error, mark 'No error' as your answer.
Hurray! Here is the toy / what I was looking for / as my birthday gift.
Option 1: Hurray! Here is the toy
Option 2: No error
Option 3: what I was looking for
Option 4: as my birthday gift.
Correct Answer: what I was looking for
Solution : The correct choice is the third option.
Explanation: The phrase "what I was looking for" is redundant and unnecessary. The correct form should simply be, "Here is the toy I was looking for". The use of "what" adds unnecessary complexity
Question : Directions: In 5 years, Rs. 5000 amounts to Rs.9000. In what time at the same rate will Rs.600 amount to Rs.900?
Option 1: 5 years
Option 2: 3 years
Option 3: 6 years
Option 4: 2 years
Correct Answer: 3 years
Solution : Given: In 5 years, Rs.5000 amounts to Rs.9000.
We have, Principal = 5000 Time = 5 yrs Amount = 9000 Rate = ? We know that, Amount (A) = P + {(P × R × T)/100} ⇒ 9000 = 5000 + {(5000 × R
Question : Select the most appropriate option that can substitute the underlined words in the following sentence.
The people who live here have had the same kind of lifestyle until hundreds of years.
Option 1: through hundreds of year
Option 2: for hundreds of years
Option 3: although hundreds of years
Option 4: since hundreds of year
Correct Answer: for hundreds of years
Solution : The correct choice is the second option.
This option correctly indicates the duration of time that the people have had the same lifestyle for hundreds of years denotes the length of time over which the lifestyle has remained unchanged.
The correct sentence
Question : Directions: In the following question, four number pairs are, given. In each pair, the number on the left side of (–) is related to the number on the right side of (–) with some Logic/Rule /Relation. Three pairs are similar based on the same Logic/Rule/Relation. Select the odd one out from the given alternatives. (NOTE: Operations should be performed on the whole numbers, without breaking down the numbers into their constituent digits. E.g.13 – operations on 13 such as adding/subtracting/multiplying etc. to 13 can be performed. Breaking down 13 into 1 and 3 and then performing mathematical operations on 1 and 3 is not allowed)
Option 1: 14 – 2762
Option 2: 16 – 4098
Option 3: 11 – 1333
Option 4: 13 – 2199
Correct Answer: 14 – 2762
Solution : Let's check each option –
First option: (14 – 2762); 143 + 2 = 2744 + 2 = 2746 ≠ 2762 Second option: (16 – 4098); 163 + 2 = 4096 + 2 = 4098 Third option: (11 – 1333); 11
Question : At how much percent above the cost price should a trader mark his goods so that after allowing a discount of 25%, he still gains 5%?
Option 1: 35
Option 2: 32
Option 3: 50
Option 4: 40
Correct Answer: 40
Solution : Let the cost price be Rs. $100$. So, the selling price is $100 + 5$ = Rs. $105$ Discount is $25$%. Let, the marked price be $x$. So, the selling price after the discount is $\frac{75}{100}×x=\frac{3x}{4}$ According to the question, $\frac{3x}{4} = 105$ $\therefore x
Question : If $\frac{x^2+1}{x}=5$, then find the value of $x^4+\frac{1}{x^4}-36$.
Option 1: 491
Option 2: 149
Option 3: 419
Option 4: 194
Correct Answer: 491
Solution : $\frac{x^2+1}{x}=5$ ⇒ $x+\frac{1}{x}=5$ Squaring both sides, we get, ⇒ $x^2+\frac{1}{x^2}+2=25$ ⇒ $x^2+\frac{1}{x^2}=23$ Squaring both sides, we get, ⇒ $x^4+\frac{1}{x^4}+2=529$ Subtracting 38 from both sides. ⇒ $x^4+\frac{1}{x^4}-36 =491$ Hence, the correct answer is 491.
Question : AB = 8 cm and CD = 6 cm are two parallel chords on the same side of the centre of a circle. If the distance between them is 2 cm, then the radius (in cm) of the circle is:
Option 1: $\frac{\sqrt{265}}{4}$
Option 2: $\frac{\sqrt{256}}{4}$
Option 3: $\frac{\sqrt{156}}{4}$
Option 4: $\frac{\sqrt{198}}{4}$
Correct Answer: $\frac{\sqrt{265}}{4}$
Solution : Given, AB = 8 cm and CD = 6 cm are two parallel chords on the same side of the centre of a circle. Let OM be perpendicular to AB and ON be perpendicular to CD. Perpendicular from the centre to a chord bisects the
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