SSC CHSL 2022 PYQs

Question 1

The value ofx2+y2,when x = 1, y = 2 is ________

Accepted Answer

Substitute x=1 and y=2 into the expression x²+y². This gives 1² + 2² = 1 + 4 = 5. Thus, the correct value is 5.

Question 2

The difference between a discount of 35% on ₹8,000 and two successive discounts of 20% and 15% on the same amount is:

Accepted Answer

Calculate the single discount equivalent to 20% and 15% using the formula A+B-(AB/100). Find the difference between this equivalent discount and 35% on ₹8,000.

Question 3

In a school, the final grades of a subject for an academic year are calculated, based on the grades that a student gets in his assignments, tests and a final exam. Assignments have the weightage of 28%, tests of 22%, and the final exam of 50%. If a student has an assignment grade of 80, a test grade of 75, an exam grade of 86, then his overall grade for the year is _________.

Accepted Answer

Calculate the weighted average: (80 * 0.28) + (75 * 0.22) + (86 * 0.50). This sums to 22.4 + 16.5 + 43 = 81.9. This is the overall grade.

Question 4

Simplify the following expression.‌17.462−4.46217.46‌+‌4.46

Accepted Answer

Use the identity a² - b² = (a - b)(a + b). The expression simplifies to (17.46 - 4.46)(17.46 + 4.46) / (17.46 + 4.46), which equals 17.46 - 4.46 = 13.

Question 5

A sells a car to B at a gain of 10% and B again sells it to C at a profit of 5%. If C pays ₹4,06,000 to B, what is the cost price of the car for A? (Consider integral part only)

Accepted Answer

Let A's cost be X. B buys at 1.1X. C buys from B at 1.05 * (1.1X) = 1.155X. Given 1.155X = 406000. So, X = 406000 / 1.155 ≈ 351515.15. Integral part is 351515.

Question 6

Simplify‌‌(r−s)‌3‌+‌‌(s−t)‌3‌+‌‌(t−r)‌36‌(r−s)‌‌(s−t)‌‌(t−r)‌

Accepted Answer

Let a = (r-s), b = (s-t), c = (t-r). Notice that a+b+c = 0. If a+b+c=0, then a³+b³+c³ = 3abc. Substitute this into the expression to get 3abc / 6abc, which simplifies to 1/2.

Question 7

If the radii of two cylinders are in the ratio of 2 ∶ 3 and their heights are in the ratio of 5 ∶ 3, then the ratio of their volumes is:

Accepted Answer

The volume of a cylinder is πr²h. For two cylinders, the ratio of volumes is (r1²/r2²) * (h1/h2). Substitute the given ratios (2/3)² * (5/3) = (4/9) * (5/3) = 20/27.

Question 8

Find the volume of a cuboid 16 m long, 12 m broad and 6 m high.

Accepted Answer

The volume of a cuboid is calculated by multiplying its length, breadth, and height. So, 16 m * 12 m * 6 m = 1152 cubic meters.

Question 9

The table below gives the numbers of laptops of three different brands sold from four different multi-brands store in the city during a certain month.Laptop BrandsDellHPLenovoStoresA304550B253240C253538D452945The number of Lenovo Laptop sold by stores A during the given month is approximately what percentage more than the number of Dell laptops sold by Store D during the same month?

Accepted Answer

Lenovo laptops from Store A = 50. Dell laptops from Store D = 45. Percentage increase = ((50 - 45) / 45) * 100 = (5/45) * 100 = (1/9) * 100 = 11.11%.

Question 10

A worker’s average income for a week is ₹180. He earned ₹115, ₹212, ₹170, ₹165, ₹185, and ₹225 from Monday to Saturday. Find his Sunday’s income.

Accepted Answer

Total income for 7 days is 7 * 180 = ₹1260. Sum of income for 6 days is 115+212+170+165+185+225 = ₹1072. Sunday's income is 1260 - 1072 = ₹188.

Question 11

X’s salary is increased by 20% and then decreased by 20%. What is the change in salary?

Accepted Answer

Use the formula for successive percentage changes: (A + B + AB/100). Here, A = +20 and B = -20. This gives -4%, indicating a 4% decrease.

Question 12

In a class, the ratio of the number of boys to that of the girls is 11 ∶ 9. 30% of the boys and 20% of the girls passed an exam. Find the percentage of students who passed the exam.

Accepted Answer

Assume 110 boys and 90 girls. Passed boys = 30% of 110 = 33. Passed girls = 20% of 90 = 18. Total passed = 33 + 18 = 51. Total students = 110 + 90 = 200. Percentage passed = (51/200) * 100 = 25.5%.

Question 13

9 ÷ [5 + 7 ÷ {9 + 9 ÷ (9 + 9 ÷ 4)}] = ?

Accepted Answer

Follow BODMAS/PEMDAS. Simplify from innermost parentheses: (9 + 9 ÷ 4) = (9 + 2.25) = 11.25. Then 9 ÷ 11.25 = 0.8. Next, {9 + 0.8} = 9.8. Then 7 ÷ 9.8 = 0.714. Finally, 9 ÷ [5 + 0.714] = 9 ÷ 5.714 ≈ 1.575. The options provided are incorrect for this question.

Question 14

What is the largest 6-digit number divisible by 52?

Accepted Answer

The largest 6-digit number is 999999. Divide 999999 by 52 to get the remainder. Subtract this remainder from 999999 to find the largest divisible number.

Question 15

The total surface area of a hemisphere is 64.57 cm2. Find its curved surface area.

Accepted Answer

Total surface area of a hemisphere is 3πr². Curved surface area is 2πr². So, curved surface area = (2/3) * Total surface area = (2/3) * 64.57 = 43.04 cm².

Question 16

The salaries of X, Y and Z are in the ratio 2 ∶ 3 ∶ 5. By what percentage is the salary of Z more than that of X?

Accepted Answer

Let salaries be 2k, 3k, 5k. Z's salary (5k) is 3k more than X's (2k). Percentage increase = (3k/2k) * 100 = 150%.

Question 17

By using faulty weight, a shopkeeper cheats to the extent of 6% while buying and selling rice. Find his gain percentage (rounded to two decimal places).

Accepted Answer

The shopkeeper gains 6% on buying and 6% on selling. The effective gain is calculated using the formula (x + y + xy/100). So, (6 + 6 + (6*6)/100) = 12 + 0.36 = 12.36%.

Question 18

A man rides his bicycle for 10 kilometres at a speed of 12 kilometres per hour, then for another 12 kilometres at a speed of 10 kilometres per hour. What is his average speed for the entire trip approximately?

Accepted Answer

Calculate total distance (10+12=22 km) and total time (10/12 + 12/10 = 0.833 + 1.2 = 2.033 hours). Average speed is total distance/total time = 22/2.033 ≈ 10.8 km/h.

Question 19

The area of a triangle whose perimeter is 36 cm and length of two sides is 17 cm and 9 cm will be:

Accepted Answer

First, find the third side using the perimeter. Then, calculate the semi-perimeter. Finally, apply Heron's formula to find the area of the triangle.

Question 20

In the following figure, ΔABC is an inscribed triangle as shown and DE is a tangent to the circle at C. Which of the angles from the given options is necessarily equal to ∠ACD?

Accepted Answer

According to the Alternate Segment Theorem, the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Thus, ∠ACD = ∠ABC.