Practice Test: Quantitative Aptitude - 13 - SSC CGL MCQ

Given: x³ + y³ = 72 and xy = 8

Solution: (x + y)³ = x³ + y³ + 3xy(x + y)

=> (x + y)³ = 72 + 3.8(x + y)

=> (x + y)³ − 24(x + y) − 72 = 0

This is a cubic equation in terms of (x + y) which has one real root = 6

=> x + y = 6

Now, (x − y)² = (x + y)² − 4xy

=> (x - y) = 2(x − y) = 2

Expression: 
Putting x = 6

=> 8 + 2p = 12

=> p = 4/2 ​= 2

Clearly, the lines x = a and y = b meets at a point (a, b)

A quadratic equation is of the form: x² − (sum of roots) x + (product of roots) = 0

Let the roots of equation are α and β

=> Sum of roots = α + β = 7

Product of roots = αβ = 12

Equation: x² − (α + β)x + αβ = 0

=> x² − 7x + 12 = 0

=> Ans - (A)

AC represents x + y = 3

BC represents 2x + 3y = 6

AB represents x = 0

=> ABC is the required triangle.

Base AB = 1 unit and height OC = 3 units

Values: 3√5​, 4√6​, 6√12​, 12√276

Taking L.C.M. of exponents, => L.C.M.(3, 4, 6, 12) = 12

Now, multiplying all the exponents by 12, we get :

Values: (5)⁴, (6)³, (12)², (276)¹

= 625, 216, 144, 276

Thus, 625 ≡ 3√5​ is the largest.

=> Ans - (A)

Given: 
Squaring both sides,

Again squaring both sides, we get:

=> 25x² − 36 = 324 + 25x² − 180x

=> 180x = 324 + 36 = 360

=> x = 360/180​ = 2

=> Ans - (C)

Substituting values from equations (ii) and (iii), we get:

x − (−x) = 2x

= ∛5​ - 7

=> Ans - (B)

Let's say one number is n and another number is p
so acc. to question sum will be 18n + 21p
and this number will be divisible by 3 so answer will be (A)

Among all options only option C has unit digit 5, and in given equation unit digit will also be 5.
So only 155 can divide the given equation completely.

If the number N is divided by 899 and leaves a remainder 65 then N = 899K + 65

and hence when N will be divided by 31

remainder of  = Remainder of 65/31 as 899 is completely divisible by 31

and hence remainder is 3

Rationalizing each term, we get, the denominator of each term will be 1, we get:

= 2 + 3 = 5

675 = 5² × 3³

and hence in order to make it a perfect cube we need to multiply it with 5.

675 x 5 = 5³ × 3³

Given: Marked Price = 50
Discount = 20%of 50 = 50 × 0.5 = 10
Hence sold price = 50 - 10 = 40
Let's say cost price is x
Profit = 25% of x (Always remember profit and loss applicable only on cost price) = x/4x
Hence sold price will be
or 5x/4 = 40
x = 32

Let marked price = 100x

After allowing discount of 15%, price = 

After further allowing discount of 10%, price = 

Now, selling price = 76.5x = 3060

=> x = 3060/76.5​ = 40

∴ Marked Price = 100 * 40 = Rs 4,000

Given Profit on selling price is 25%
Suppose selling price is y
hence profit will be y/4y and cost price will be 3y/4
Now profit percentage on cost price will be

i.e. 100/3​ = 33.33

B's capital be x.

x = 18000

Let the printed price be 100x and the cost price be 100y

After discount of 10%, selling price of the book = = 90x
After earning a profit of 12%, selling price = 
= 112y

Now, equating above equations, we get:

=> 90x = 112y

T₃​ = a + 2d = 13-------(1)

T₅​ = a + 4d = 21-------(2)

on solving (1) AND (2)

d = 4 & a = 5

T_13​ = a + 12d = 5 + 12(4) = 5 + 48 = 53

So the answer is option A.

First term of AP = a = -19 and last term = l = 36

Number of terms = n = 12

= 17 × 6 = 102

=> Ans - (C)

Let the first term of an AP = a and the common difference = d

3th term of AP = A₃​ = a + 2d = −13 ----------(i)

8th term = A_8​ = a + 7d = 2 --------(ii)

Subtracting equation (i) from (ii), we get :

=> 7d − 2d = 2 − (−13)

=> 5d = 15

=> d = 15/5 ​= 3

Substituting it in equation (ii), => a = 2 − 7(3) = 2 − 21 = −19

∴ 14th term = A₁₄​ = a + 13d

= −19 + 13(3) = −19 + 39 = 20

=> Ans - (C)

Let the first term of an AP = a and the common difference = d

4th term of AP = A₄​ = a + 3d = 11 ----------(i)

7th term = A7 ​= a + 6d = −4 --------(ii)

Subtracting equation (i) from (ii), we get:

=> 6d − 3d = −4 − 11

=> 3d = −15

=> d = −15/3 ​= −5

Substituting it in equation (i), => a = 11 − 3(−5) = 11 + 15 = 26

∴ 15th term = A₁₅ ​= a + 14d

= 26 + 14(−5) = 26 − 70 = −44

=> Ans - (B)