SSC GD 2021 PYQs

Question 1

Find the value of 6 ÷ [4 × 3 {5 × 4 ÷ (5 of 4 + 4 ÷ 2 × 4)}].

Accepted Answer

Apply BODMAS rule: first solve 'of', then division, multiplication, addition, and finally the outer operations. The expression simplifies to 6 ÷ [4 × 3 {5 × 4 ÷ (20 + 8)}] = 6 ÷ [12 {20 ÷ 28}] = 6 ÷ [12 × 5/7] = 6 ÷ 60/7 = 6 × 7/60 = 7/10.

Question 2

The curved surface area of a right circular cylinder of height 14 cm is 176 cm2. What is the area (in cm2) of the base of the cylinder? (Take π =‌227)

Accepted Answer

Given CSA = 2πrh = 176 cm² and h = 14 cm. Substitute values to find r = 2 cm. Base area = πr² = (22/7) * 2² = 88/7 cm².

Question 3

P, Q and R can complete a work in 10 days, 20 days and 30 days, respectively, working alone. How soon can the work be completed if P is assisted by Q and R on alternate days?

Accepted Answer

P's 1-day work = 1/10. Q's = 1/20. R's = 1/30. Day 1: P+Q work = 1/10+1/20 = 3/20. Day 2: P+R work = 1/10+1/30 = 4/30. Total work in 2 days = 3/20+4/30 = 17/60. Work in 6 days = 3 * 17/60 = 51/60. Remaining work = 9/60. On Day 7, P+Q do 3/20 = 9/60 work. So, total 7 days.

Question 4

If the sides of the triangle measure 21 cm, 35 cm, and 28 cm. What is the measure of its in radius? (in cm)

Accepted Answer

The sides 21, 28, 35 form a right-angled triangle (3k, 4k, 5k where k=7). The inradius (r) of a right triangle is (a+b-c)/2. So, r = (21+28-35)/2 = 14/2 = 7 cm.

Question 5

If the compound interest on a certain sum of 2 years at 4% per annum is ₹1530. What would be the simple interest on the same sum for the same period and at the same rate?

Accepted Answer

For 2 years, CI - SI = P(R/100)^2. Also, CI = SI + SI*(R/200). Given CI = 1530, R = 4%. So, 1530 = SI + SI*(4/200) = SI + SI/50 = 51SI/50. Thus, SI = 1530 * 50 / 51 = 30 * 50 = ₹1500.

Question 6

If 25% of x is 40 less than 30% of (x + 60), then 35% of (x- 40) is what percentage more than 120?

Accepted Answer

Set up the equation: 0.25x = 0.30(x + 60) - 40. Solve for x to get x = 440. Calculate 35% of (440 - 40) = 35% of 400 = 140. The percentage more than 120 is ((140 - 120) / 120) * 100 = 16.66%.

Question 7

The least number of square tiles required to pave the floor of a room with dimensions 15 meters and 12 meters will be:

Accepted Answer

Find the HCF of 15 and 12, which is 3 meters. This is the side length of the largest square tile. The number of tiles is (room area) / (tile area) = (15*12) / (3*3) = 180/9 = 20.

Question 8

By selling 40 dozen bananas for ₹ 1,440, a man gains 20%. In order to gain 30%, for how much should he sell 20 dozen bananas ?

Accepted Answer

First, find the cost price of 40 dozen bananas. Then, calculate the selling price for 20 dozen bananas to achieve a 30% profit. The selling price for 20 dozen is ₹780.

Question 9

A single discount equivalent to three successive discounts i.e. 7%, 12% and 5% is:

Accepted Answer

Calculate the effective discount for 7% and 12% first using A+B-(AB/100). Then, apply this result with the 5% discount using the same formula to find the single equivalent discount.

Question 10

Simplify the following expression.	ext"$\;{\;{3}/{4}\;of\;	ext\;""\;"\;1{3}/{5}−1\;×\;5\;+\;2}/{24−6\;+\;2} ÷ 4$

Accepted Answer

Using BODMAS: Numerator = 3/4 of 8/5 - 5 + 2 = 6/5 - 3 = -9/5. Denominator = 24 - 6 + 2 = 20. Result = (-9/5) / 20 ÷ 4 = -9/400.

Question 11

Two numbers are 8% and 28%, respectively, more than a third number. What is the ratio between the two numbers?

Accepted Answer

Assume the third number is 100. The first number is 108 (100+8%) and the second is 128 (100+28%). The ratio is 108:128, which simplifies to 27:32.

Question 12

An article was sold at‌118times of its cost price. The profit per cent is:

Accepted Answer

If SP = (11/8)CP, then Profit = SP - CP = (11/8)CP - CP = (3/8)CP. Profit percentage is (Profit/CP) * 100 = (3/8) * 100 = 37.5%.

Question 13

Find the ratio of the diagonal of a square of side 15 cm, to its side.

Accepted Answer

The diagonal of a square with side 'a' is a√2. The ratio of diagonal to side is a√2 : a, which simplifies to √2 : 1.

Question 14

Radha saves 28% of her income. If her expenditure increases by 25% and savings increase by 30%, then by what percentage will her income increase?

Accepted Answer

Assume income is 100. Savings = 28, Expenditure = 72. New expenditure = 72 * 1.25 = 90. New savings = 28 * 1.30 = 36.4. New income = 90 + 36.4 = 126.4. Percentage increase = (126.4 - 100)/100 * 100 = 26.4%.

Question 15

A batsman scored 124 runs, which included 6 boundaries and 10 sixes. What percentage of his total score did he make by running between the wickets?

Accepted Answer

Runs from boundaries (6*4) = 24. Runs from sixes (10*6) = 60. Total runs from boundaries/sixes = 84. Runs by running = 124 - 84 = 40. Percentage = (40/124) * 100 = 32.25% or 32 8/31%.

Question 16

Two trains start from places A and B. respectively, and travel towards each other at the speeds of 60 km/h and 50 km/h. respectively. By the time they meet, the faster train has travelled 110 km more than the slower train. What is the distance between A and B?

Accepted Answer

Let time be T. Faster train travels 60T, slower 50T. Difference is 10T = 110 km, so T = 11 hours. Total distance = (60+50) * 11 = 110 * 11 = 1210 km.

Question 17

By selling 18 oranges for ₹151.20, a man loses 20%. How many oranges should he sell for ₹819 to earn a profit of 30%?

Accepted Answer

First, find the cost price of one orange. Then, calculate the selling price per orange for a 30% profit. Finally, divide the target amount by this new selling price to find the number of oranges.

Question 18

A man bought two mobiles phones for ₹50,000. He sold one of them at a profit of 15% and the other at a loss of 20%. If the selling price of each mobile is the same, what is the approximate cost price (to the nearest rupee) of the mobile that was sold at a loss?

Accepted Answer

Let CP1 and CP2 be the cost prices. SP1 = 1.15 * CP1 and SP2 = 0.80 * CP2. Since SP1 = SP2, 1.15 * CP1 = 0.80 * CP2. Also, CP1 + CP2 = 50000. Solve these equations to find CP2 ≈ ₹29,487.

Question 19

The value of[6+3 of\;	ext""‌7−5]÷2 of\;	ext""‌11is:

Accepted Answer

Apply BODMAS: first 'of' (multiplication), then addition/subtraction inside brackets, finally division. [6 + 21 - 5] ÷ 22 = [27 - 5] ÷ 22 = 22 ÷ 22 = 1.

Question 20

The amount obtained by investing a certain sum at r% p.a. for 3 years at simple interest is equal to the simple interest on the same sum at the same rate for 19 years. What is the value of r ?

Accepted Answer

Let the sum be P. Amount = P + (P*r*3)/100. Simple Interest = (P*r*19)/100. Equating them: P + 3Pr/100 = 19Pr/100. This simplifies to 1 = 16r/100, so r = 100/16 = 6.25.