Quantitative Aptitude - 15 - Free MCQ Practice Test with solutions, SSC

Start from the root of 169 then second root will reduce to 11, third root will reduce to 7, fourth root will reduce to 8, and finally, it reduce to value 4.

Put a = 5 and b = 3 in given equation
hence it will be 5³ = 125

In an arithmetic progression with first term, a = -23, last term, l = 42

Number of terms = n = 14

7 × 19 = 133

=> Ans - (B)

Product of: 6ab, 3b and 4a

= (6 × 3 × 4) × (ab × b × a)

= 72a²b²

=> Ans - (A)

Expression: 2714/5074

HCF of 2714 and 5074 = 118
Dividing numerator and denominator by 118

= 23/43

=> Ans - (C)

Average of 5 numbers is given as 56.4

Total of 5 numbers = Average x Number of Elements

Total of 5 numbers = 56.4 x 5 = 282

Average of 1st and 2nd number is 53

Total of 1st and 2nd number = 53 x 2 = 106

Average of 4th and 5th number is 56

total of 4th and 5th number = 56 x 2 = 112

1st + 2nd + 3rd + 4th + 5th = 282

106 + 3rd + 112 = 282

3rd = 64

hence option A is correct

Let's say marked price = 100
Then cost price = 64
 and selling price = 88
discount = 24

Let's assume no. of seats were 100 and price of a ticket was 100 hence total revenue is 10000.
Now acc. to question after increment no. of seats will be 125 and price of a ticket will be 112. Hence total revenue will be 125 * 112 = 14000.
Increment in revenue = 4000
% increase in revenue = 40%

Let's say total number of workers are xx.
Hence above 30 years workers will be 60x/100​ 
And male workers will be 
So x will be = 3000.

Let the total capital be Rs z and total time be y

hence ratio of profit divison for A, B, C will be

A : B : C = 4 : 9 : 84

hence profit for B = 9/97 x 19400 = 1800

Number of subjects = 6

wrong average = 88

sum of marks calculated wrongly in all subjects = 88 x 6 = 528

Correct Marks = 528 - 86 + 68 = 510

Correct Average of marks = 510/6​ = 85

Let the amount with C be Rs y

so, amount with B = y +6

and amount with A = y + 13

and it is given that total amount is 76, So

y + y + 6 + y + 13 = 76

y= 19

Hence A : B : C = 32 : 25 : 19

cost of 5 kg. rice = 100 rs.
cost of 1 kg. rice = 20 rs.
Before discount cost price will be = 

Total share percentage on food and rent = 45 + 14 = 59%
Hence price will be = 25000 × 59/100 = 14750

Marked price(M) of article = Rs 975

Selling Price (S) of article = Rs 897

= 8%

→ 24 = 20 + 4x

→ x = 1

so the answer is option D.

Reflection of point (x, y) in line x = a is (-x + 2a, y)

Now, Reflection of point (-2, 5) in line x = -1

= [2 + 2(-1), 5] = (0, 5)

=> Ans - (B)

Slope of line passing through points (4, -2) and (-3, 5)

Slope of two parallel lines is always equal.

=> Slope of the line parallel to the line having slope -1 = -1

=> Ans - (D)

Reflection of point (x, y) in line x = a is (-x + 2a, y)

Now, Reflection of point (-3, 1) in line x = -2

= [3 + 2(-2), 1] = (-1, 1)

=> Ans - (A)

(x, y) = (4, 5)

So the answer is option B.

Production of C in year 2003-2004 = 6 per 100 kg in lakh
in year 2004-05 = 11 per 100 kg in lakh
in year 2005-06 = 15 per 100 kg in lakh
Total will be 32 per 100 kg in lakh or 320000000 kg

Total no. of electronic items in:

2009 = 6000 + 7000 = 13000

2010 = 9000 + 9400 = 18400

2011 = 13000 + 9000 = 22000 [MAX]

2012 = 11000 + 10000 = 21000

2013 = 8000 + 12000 = 20000

=> Total production is maximum in 2011

Production of Company A = 1400 lakh tonnes

Demand of company C = 2600 lakh tonnes

Production of company A as a percentage of demand of company C = 1450/2600 = 55% (Approx)

Option D is the right answer.

Total demand = 2100 + 3150 + 2600 + 5000 + 2800 + 3300 = 18950

Total supply = 1450 + 3650 + 3100 + 4200 + 3700 + 4500 = 20600

Difference in total demand and supply = 1650

Average difference = 1650/6 = 275

Hence, option C is the right answer.

In 2000, no. of students who passed with:

1st class = 90

2nd class = 50

3rd class = 20

Total students = 90 + 50 + 20 = 160

=> Required ratio = 50/160​ = 5 : 16