Quantitative Aptitude - 11 - Free MCQ Practice Test with solutions, SSC

Degree of expenditure on food = 120

Degree of expenditure on savings = 60

=> Required ratio = 120/6 = 2 : 1

In 2002, no. of students who passed in :

1st class = 120

2nd class = 60

3rd class = 15

=> Total no. of students appeared for class 10th exam in 2002 = 120 + 60 + 15 = 195

Expenditure in 2002 = 30

Expenditure in 2003 = 40

So, the answer is option A.

In shop A, percentage of violet balls sold as a percentage of total number of balls = 35%

Percentage of yellow balls sold as a percentage of total number of balls = 20%

Ratio = 35 : 20 = 7 : 4

Option B is the right answer.

Only in A and C there is a steady increment in production of cotton as in D and E, It is decreased and in B production is equal for two years. Hence answer will be C.

Multiplying P inside whole equation will reduced to ((P + 1)³)^1/3​ = (p + 1)
hence answer will be 125

as x + y = 4
and xy = 1
Now after putting values of x + y and xy and solving, we will get its value as 7/26.

a² + b² + 4c² = 2(a + b − 2c − 3)
So, (a − 1)² + (b − 1)² + (2c + 1)² = 0
Or, a = 1, b = 1, c = −1​/2
So, a² + b² + c² = 1 + 1 + 1/4 ​= 9/4

Given x² − 3x + 1 = 0
or x² − 3x = −1
or (x − 3) = −1/x

Now according to given question

After putting value of x + 1/x in above equation, it will get reduced to 10.

Let the number be x

According to ques,

=> 7x² + 21 = 52x

=> 7x² − 52x + 21 = 0

=> 7x² − 49x − 3x + 21 = 0

=> 7x(x - 7) - 3(x - 7) = 0

=> (7x - 3)(x - 7) = 0

=> x = 7, 3/7

=> Ans - (C)

As we know (a − b)² = a² + b² − 2ab
We assume that first number is a and second number is b hence ab = 45
 and a - b = 4
after putting values we will get a² + b² = 106

Given: ∠B = 90 and AB : BC = 2 : 1

To find: sin A + cot C = ?

Solution: Let AB = 2x and BC = x

In right △ABC


Now, sin A + cot C

Given: ∠A = 60° and AB = 12 cm

To find: BD = ?

Solution: Since, all the sides of a rhombus are equal, => AB = AD

=> ∠ABD = ∠ADB

Now, in △ABD

=> ∠ABD + ∠ADB + ∠A = 180°

=> 2∠ABD = 120°

=> ∠ABD = ∠ADB = ∠A = 60°

=> ABD is equilateral triangle.

=> AB = AD = BD = 12 cm

Maximum number of intersection points of n lines = ​

Thus, in a plane of 4 lines, maximum intersection points = 

=> 7 is not possible.

=> Ans - (D)

Given question can be written as 9 + 5 + 6√5​ it is 3² + √5² + 2(3)(√5) = a² + b² +2ab
or it will be square of 3 + √5​

Difference between compound interest and simple interest for 2 years will be 
= (where P is principal amount 5000 and r is rate )
after solving above equation we will get r = 8%

It is given that Rs 6000 with interest 4% per annum (SI) produces an interest in 5 years = 

= Rs 1200

Now the new principal amount is Rs 8000 and rate of interest is 3 % per annum. Let the time be T years in which same Rs 1200 will be generated as interest.

T = 5 years

We know that:

1. For first year, compound interest and simple interest is same if the principal amount and rate of interest is same in both cases.

2. From 2nd year onwards, the compound interest is normal interest plus the interest on accumulated amount due to interest until last cycle.

3. Every year simple interest remains same if Rate of Interest and principal amount remains same.

Let the compound interest for 1st year be Rs y

For two years, CI = Rs 410

y = 200

So for two years , Simple Interest = 200 + 200 = Rs 400

let the sum of money be Rs P

And rate of interest = R %

It is given that for two years the Compound Interest = Rs 2050

As it is given that for 3 years the Simple Interest = Rs 3000 and we know that SI for every year is same so for 1st year

Simple interest = Rs 1000

For 1st year SI and CI are same if rate of interest and Principal amount is same and hence for 1st year CI = 1000

For 2 years the CI = 2050 = 1000 + 1000 + R/100 x 1000)

R = 5%

P = Rs 20000

Let the given sum = Rs. 100x

Rate of interest = 18% and time period = 2 years


=> Difference between simple and compound interests = 

∴ Value of given sum = 100 × 25 = Rs.2500

Given:
Numbers: First = 24
               Second = x (suppose)
               H.C.F. of numbers = 8
               L.C.M. of numbers = 48
As we know:
H.C.F. * L.C.M. = Product of numbers
Hence
48 * 8 = 24 * x
x = 16

We assume that numbers are hr₁​ and hr_2​ (where h = H.C.F. of numbers and r₁​ and r_2​ are prime factors)
So L.C.M. will be = hr_1​r_2​ = 120
or r₁​r₂​ = 12
So r₁​ = 4 and r_2​ = 3; numbers will be 40 and 30, sum is 70
or r₁​ = 12 and r₂​ = 1; numbers will be 120 and 10, sum is 130
Hence only option D justifies.

Let the numbers be 3x, 4x
LCM of 3x and 4x is = 12x
So the number 84 is divisible by 12
84/12​ = 7
The numbers are 7x3 = 21 , 7x 4 = 28
The greatest number is 28

Factors of 57 = 1, 3, 19, 57

Factors of 513 = 1, 3, 9, 19, 27, 57, 171, 513

The common factors are = 1, 3, 19, 57

=> Highest common factor = 57

=> Ans - (B)

Let's say numbers are x and y
hence sum of the reciprocals will be 1/x​ + 1/y​
or
as x + y = 36 (given)
and xy = HCF × LCM
                  = 3 × 105 = 315
after putting the values we will get summation of reciprocals equals to 4/35