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Practice Test: Quantitative Aptitude - 12 - SSC CGL MCQ


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25 Questions MCQ Test - Practice Test: Quantitative Aptitude - 12

Practice Test: Quantitative Aptitude - 12 for SSC CGL 2024 is part of SSC CGL preparation. The Practice Test: Quantitative Aptitude - 12 questions and answers have been prepared according to the SSC CGL exam syllabus.The Practice Test: Quantitative Aptitude - 12 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Quantitative Aptitude - 12 below.
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Practice Test: Quantitative Aptitude - 12 - Question 1

An employer reduces the number of employees in the ratio 8 : 5 and increases their wages in the ratio 7 : 9. As a result, the overall wages bill is

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 1

Let's say employees were change from 8x to 5x and wage per employee is changed from 7y to 9y
Hence total wage is changed from 56xy to 45xy or in a ratio of 56 : 45

Practice Test: Quantitative Aptitude - 12 - Question 2

The ratio of ages of father and son is 7 : 2. Five years ago the product of their ages was 150. What is the age of the father?

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 2

Let father's age = 7x years and son's age = 2x years

According to ques, => (7x − 5)(2x − 5) = 150

=> 14x2 − 35x − 10x + 25 − 150 = 0

=> 14x2 − 45x − 125 = 0

=> 14x2 − 70x + 25x − 125 = 0

=> 14x(x − 5) + 25(x − 5) = 0

=> (x − 5)(14x + 25) = 0

Since, age can't be negative, thus x = 5

∴ Father's age = 7 × 5 = 35 years

=> Ans - (C)

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Practice Test: Quantitative Aptitude - 12 - Question 3

The sum of the ages of husband and wife at present is 56. Ten years ago the product of their ages was 320. What is the age of the husband and the wife?

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 3

Let husband's age = xx years

=> Wife's age = (56 − x) years

According to ques, (x − 10)(56 − x − 10) = 320

=> (x − 10)(46 − x) = 320

=> 46x − x2 − 460 + 10x = 320

=> x2 − 56x + 780 = 0

=> x2 − 30x − 26x + 780 = 0

=> x(x − 30) − 26(x − 30) = 0

=> (x − 30)(x − 26) = 0

=> x = 30, 26

∴ Ages of the husband and the wife = 30 and 26 years.

Practice Test: Quantitative Aptitude - 12 - Question 4

Ronald and Elan are working on an Assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much time will they take working together on two different computers to type an assignment of 110 pages?

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 4

Ronald takes 6 hours to type 32 pages

no.of pages typed by ronald per hour = 32/6 = 16/3

 Elan takes 5 hours to type 40 pages

no.of pages typed by elan per hour = 40/5 = 8

in 1hr, both can type 8 + 16/3 pages = 40/3

time taken by both to type 110 pages = 110/(40/3) = 33/4 = 8hrs 15mins    (∵ 1/4 hr = 15mins)

so the answer is option C.

Practice Test: Quantitative Aptitude - 12 - Question 5

The ratio of present ages of Ramya and Saurabh is 8 : 7. After 10 years the ratio of their ages will be 12 : 11. What is Ramya's present age?

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 5

Let Ramya's present age = 8x years and Saurabh's present age = 7x years

According to ques,

=> 88x + 110 = 84x + 120

=> 88x - 84x = 120 - 110 = 10

=> x = 10/4 ​= 2.5

∴ Ramya's age = 8 × 2.5 = 20 years

=> Ans - (A)

Practice Test: Quantitative Aptitude - 12 - Question 6

If a + 1/a + 2 = 0, then the value of

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 6

It is given that 
it is possible only when a = -1
hence a + 2 = 1
and so

Practice Test: Quantitative Aptitude - 12 - Question 7

If 5x + 9y = 5 and 125x3 + 729y3 = 120 then the value of the product of x and y is

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 7

Expression: 5x + 9y = 5

Cubing both sides, we get:

=> (5x + 9y)3 = 125

=> 125x3 + 729y3 + 135xy(5x + 9y) = 125

=> 125x3 + 729y3 + 135xy ∗ 5 = 125

Since, 125x3 + 729y3 = 120

Practice Test: Quantitative Aptitude - 12 - Question 8

The value of

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 8


 

on squaring both sides

x2 − x − 6 = 0

x = 3 , x = -2

here -2 will be rejected as square root can not give negative value and hence x = 3

Practice Test: Quantitative Aptitude - 12 - Question 9

If x = 2 + √3 , then the value,

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 9

x = 2 + √3​

1/x ​= 2 − √3​

so the answer is option B.

Practice Test: Quantitative Aptitude - 12 - Question 10

If m - 5n = 2, then the value of (m3 − 125n3 - 30 mn) is

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 10

Using the formula, (x − y)3 = x3 − y3 − 3xy(x − y)

=> (m − 5n)3 = m3 − 125n3 − 15mn(m − 5n)

=> 23 = m3 − 125n3 − 15mn ∗ 2

=> m3 − 125n3 − 30mn = 8

Practice Test: Quantitative Aptitude - 12 - Question 11

Which of the following statement (s) is / are TRUE?

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 11


L.H.S. = 26 + 2.6 + 0.26 = 28.86 ≠ R.H.S.

Thus, only II is correct.

=> Ans - (B)

Practice Test: Quantitative Aptitude - 12 - Question 12

The 4th term of an arithmetic progression is 15, 15th term is -29, find the 10th term?

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 12

The nth term of an A.P. = a + (n - 1)d, where 'a' is the first term, 'n' is the number of terms and 'd' is the common difference.

4th term, A4 = a + (4 - 1)d = 15

=> a + 3d = 15 -----------------(i)

Similarly, 15th term, A15 = a + 14d = -29 ------------------(ii)

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

=> (14d - 3d) = -29 - 15

=> d = -44/11 ​= −4

Substituting it in equation (i), => a - 12 = 15

=> a = 15 + 12 = 27

∴ 10th term, A10 = a + (10 - 1)d

= 27 + (9× − 4) = 27 − 36 = −9

=> Ans - (D)

Practice Test: Quantitative Aptitude - 12 - Question 13

If 4x = 71024​, then what is the value of x?

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 13

 

Expression: 4x = 7√1024

=> Ans - (A)

Practice Test: Quantitative Aptitude - 12 - Question 14

What is the simplified value of

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 14

Expression:

=> Ans - (B)

Practice Test: Quantitative Aptitude - 12 - Question 15

What is the value of

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 15

Expression: 

= 8 + 13 + 6 + 15 = 42

=> Ans - (C)

Practice Test: Quantitative Aptitude - 12 - Question 16

If the cost price of 15 articles is equal to the selling price of 12 articles, find gain %

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 16

Let's say cost price of 15 articles is xx
cost price of 12 articles will be = 12x​/15
selling price of 15 articles is = x
gain = 3x​/15

Practice Test: Quantitative Aptitude - 12 - Question 17

A man bought an article listed at Rs. 1500 with a discount of 20% offered on the list price. What additional discount must be offered to man to bring the net price to Rs. 1,104?

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 17

After having 20% discount price will be =
So for net price of 1104 discount should be 1200 - 1104 = 96

Practice Test: Quantitative Aptitude - 12 - Question 18

A is 50% as efficient as B. C does half of the work done by A and B together. If C alone does the work in 20 days, then A, B and C together can do the work in

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 18

C does 1/20​ amount of work in a day.
And A + B do twice of work as much done by C or A + B = 2C or 2/20​.
So A + B + C together will do amount of work in a day.
And complete work will be done in 20/3​ days.

Practice Test: Quantitative Aptitude - 12 - Question 19

A man sold 20 apples for Rs. 1000 and gained 20%. How many apples did he buy for Rs. 1000?

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 19

Selling price of 20 apples = 1000
gain = 20% 
Cost price of 20 will be = 
So on cost price of 1000 amount of apples will be 

Practice Test: Quantitative Aptitude - 12 - Question 20

A grain dealer cheats to the extent of 10% while buying as well as selling by using false weights. His total profit percentage is:

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 20

Let the retailer buys 110 gram in Rs 100 because of 10 % cheating 

so cost price = 1 gram = Rs 100/110 = Rs 0.909

and retailer sells 90 gram in Rs 100 

and hence selling price = Rs 100/90 = Rs 1.111

Practice Test: Quantitative Aptitude - 12 - Question 21

What is the equation of the line whose y­intercept is ­3/4 and making an angle of 45° with the positive x-­axis?

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 21

Slope of line making an angle of 45° with the positive x­-axis = tan(45°)

=> Slope, m = 1

y-intercept, c = 3/43​

Equation of line having slope mm and y intercept cc is: y = mx + c

=> y = x + 43​

=> 4y = 4x + 3

=> 4x − 4y = −3

=> Ans - (B)

Practice Test: Quantitative Aptitude - 12 - Question 22

In what ratio is the segment joining (-1, -12) and (3, 4) divided by the x-axis?

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 22

Using section formula, the coordinates of point that divides line joining A = (x1​, y1​) and B = (x2​, y2​) in the ratio a : b

Let the ratio in which the segment joining (-1,-12) and (3,4) divided by the x-axis = k : 1

Since, the line segment is divided by x-axis, thus y-coordinate of the point will be zero, let the point of intersection = (x, 0)

Now, point P (x, 0) divides (-1, -12) and (3, 4) in ratio = k : 1

=> 4k - 12 = 0

=> k = 12​/4 = 3

∴ Required ratio = 3 : 1

=> Ans - (C)

Practice Test: Quantitative Aptitude - 12 - Question 23

Find equation of the perpendicular bisector of segment joining the points (2, -5) and (0, 7)?

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 23

Let line l perpendicularly bisects line joining  A(2, -5) and B(0, 7) at C, thus C is the mid point of AB.

=> Coordinates of C = 

Now, slope of AB 

Let slope of line l = m

Product of slopes of two perpendicular lines = -1

=> m × −6 = −1

=> m = 1/61

Equation of a line passing through point (x1​, y1​) and having slope mm is (y − y1​) = m(x − x1​)

∴ Equation of line l

=> (y - 1) = 1/6​(x − 1)

=> 6y - 6 = x - 1

=> x - 6y = 1 - 6 = -5

=> Ans - (C)

Practice Test: Quantitative Aptitude - 12 - Question 24

Find k, if the line 2x - ­3y = 11 is perpendicular to the line 3x + ky = ­4?

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 24

Slope of line 2x - ­3y = 11 is
= 2/3
Slope of line 3x + ky = ­4 is −3​/k

Also, product of slopes of two perpendicular lines is -1

=> k = −2 × −1 = 2

=> Ans - (D)

Practice Test: Quantitative Aptitude - 12 - Question 25

The point P(-4, 1) divides the segment joining the points (x, 0) and (0, y) in the ratio 3 : 2. Find x and y?

Detailed Solution for Practice Test: Quantitative Aptitude - 12 - Question 25

Using section formula, the coordinates of point that divides line joining A = (x1​, y1​) and B = (x2​, y2​) in the ratio a : b

Now, point P (-4, 1) divides (x, 0) and (0, y) in ratio = 3 : 2

=> 3y = 5 => y = 5/3

=> Ans - (A)

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