Test: Linear Equations - 2 - SSC CGL MCQ

# Test: Linear Equations - 2 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - Test: Linear Equations - 2

Test: Linear Equations - 2 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The Test: Linear Equations - 2 questions and answers have been prepared according to the SSC CGL exam syllabus.The Test: Linear Equations - 2 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Linear Equations - 2 below.
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Test: Linear Equations - 2 - Question 1

### The difference between the squares of two numbers is 256000 and the sum of the numbers is 1000. The numbers are

Detailed Solution for Test: Linear Equations - 2 - Question 1

Let us assume the numbers are a and b.
According to question,
a2 - b2 = 256000 .......................(1)
and a + b = 1000 .........................(2)
On dividing the Equation (1) with Equation (2),
(a2 - b2)/(a + b) = 256000/1000
(a2 - b2)/(a + b) = 256
(a + b)(a - b)/(a + b) = 256
a - b = 256...............................(3)
Add the equation (2) and (3), we will get
a + b + a - b = 1000 + 256
2a = 1256
a = 628
Put the value of a in equation (2), we will get
658 + b = 1000
b = 1000 - 628
b = 372
So answer is 628 , 372.

Test: Linear Equations - 2 - Question 2

### The numbers are such that the square of one is 224 less than 8 times the square of the other. If the numbers be in the ratio 3 : 4, the numbers are

Detailed Solution for Test: Linear Equations - 2 - Question 2

Let us assume the ratio term is r.
then the number will be 3r and 4r.
According to question,
(4r)2 + 224 = 8 x (3r)2
16r2 + 224 = 8 x 9r2
16r2 + 224 = 72r2
72r2 - 16r2 = 224
56r2 = 224
r2 = 224/56
r2 = 4
r = 2
First number = 3r = 3 x 2 = 6
Second number = 4r = 4 x 2 = 8

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Test: Linear Equations - 2 - Question 3

### The sum of three consecutive odd numbers is 20 more than the first of these numbers.What is the middle number ?

Detailed Solution for Test: Linear Equations - 2 - Question 3

Let us assume the first Odd number is a.
then 2nd odd number = a + 2
and 3rd odd number = a + 4
According to question,
Sum of three consecutive odd numbers is 20 more than the first of these numbers;
a + (a + 2) + (a + 4) = a + 20
⇒ a + a + 2 + a + 4 - a = 20
⇒ 2a + 6 = 20
⇒ 2a + 6 = 20
⇒ 2a = 20 - 6
⇒ 2a = 14
⇒ a = 7
The First odd number a = 7
The second odd number = a + 2 = 7 + 2 = 9
Second odd number is middle number = 9

Test: Linear Equations - 2 - Question 4

Two-fifths of one -fourth of three -seventh of a number is 15. What is half of that number?

Detailed Solution for Test: Linear Equations - 2 - Question 4

Let us assume the number is P.
According the question,
three -seventh of a number = 3/7 x P
one -fourth of three -seventh of a number = 1/4 x 3/7 x P
Two-fifths of one -fourth of three -seventh of a number = 2/5 x 1/4 x 3/7 x P
⇒ 2/5 x 1/4 x 3/7 x P = 15
⇒ 2 x 1 x 3 x P = 15 x 7 x 4 x 5
⇒ P = 15 x 7 x 4 x 5 / 2 x 3
⇒ P = 5 x 7 x 2 x 5
⇒ P = 350
∴ P /2 = 350/2
⇒ P /2 = 175

Test: Linear Equations - 2 - Question 5

A person on tour has total ₹360 for his daily expenses. He decides to extend his tour programme by 4 days which leads to cutting down daily expenses by ₹3 a day. The numbers of days of tour are :

Detailed Solution for Test: Linear Equations - 2 - Question 5

Let us assume the numbers of days of tour before extending the tour are D and daily expenses is E.
According to question,
Total expenses on tour = per day expenses x total number of days
Total expenses on tour = E x D = ED
360 = ED
⇒ ED = 360
⇒ E = 360/D .................(1)
Again according to question,
After extending the tour by 4 days, Number of days of tour = D + 4
Expenses will be reduced by 3 rupees, then everyday expenses = E - 3
so total expenses = (D + 4) x (E - 3)
360 = (D + 4) x (E - 3)
(D + 4) x (E - 3) = 360 .............................................(2)
Put the value of E in Equation (2). we will get,
(D + 4) x (360/D - 3) = 360
⇒ (D + 4) x ( (360 - 3D)/D ) = 360
⇒ (D + 4) x ( (360 - 3D) ) = 360 x D
⇒ (D + 4) x (360 - 3D) = 360 x D
After multiplication by algebra law,
⇒ 360 x D - 3D x D + 4 x 360 - 4 x 3D = 360D
⇒ 360 x D - 3D + 1440 - 12D = 360D
⇒ - 3D + 1440 - 12D = 360D - 360 x D
⇒ - 3D + 1440 - 12D = 0
⇒ 3D - 1440 + 12D = 0
⇒ D - 480 + 4D = 0
⇒ D + 4D - 480 = 0
⇒ D + 24D - 20D - 480 = 0
⇒ D(D + 24) - 20(D + 24) = 0
⇒ (D + 24) (D - 20) = 0
Either (D + 24) = 0 or (D - 20) = 0
So D = - 24 or D = 20
But days cannot be negative so D = 20 days.

Test: Linear Equations - 2 - Question 6

The auto-rickshaw fare consists of a fixed charge together with the charge for the distance covered. For a journey of 10 km, the charge paid is ₹85 and for a journey of 15 km, the charge paid is ₹120. The fare for a journey of 25 km will be

Detailed Solution for Test: Linear Equations - 2 - Question 6

Let use assume the fixed charge = ₹ a
and charge for 1 km is ₹ = b
According to question,
for 10 KM journey charge paid = 85
a + 10 x b = 85
a + 10b = 85 .........................(1)
for 15 KM journey charge paid = 120
a + 15b = 120.........................(2)
Subtract the equation (1) from equation (2). we will get,
a + 15b - a - 10b = 120 - 85
5b = 35
b = 7
Put the value of b in equation (1). we will get
a + 10 x 7 = 85
a = 85 - 70
a = 15
Charges for 25 km = a + 25 x b
Put the value of a and b in above equation.
Charges for 25 km =15 + 25 x 7 = 15 + 175 = 190
Charges for 25 km =₹190

Test: Linear Equations - 2 - Question 7

The sum of three consecutive multiples of 3 is 72. What is the largest number ?

Detailed Solution for Test: Linear Equations - 2 - Question 7

Let us assume the number be 3p , 3(p+1) and 3(p+2)
According to question,
3p + 3(p+1) + 3(p+2) = 72
⇒ 3p + 3p + 3 +3p + 6 = 72
⇒ 9p +9 = 72
⇒ 9p = 72 - 9
⇒ 9p = 63
⇒ p = 63/9 = 7
∴ Largest number = 3(p + 2)
Put the value of p in above equation.
⇒ Largest number = 3 x (7 + 2)
⇒ Largest number = 3 x 9
⇒ Largest number = 27

Test: Linear Equations - 2 - Question 8

The product of two numbers is 192 and the sum of these two numbers is 28. What is the smaller of these two numbers ?

Detailed Solution for Test: Linear Equations - 2 - Question 8

Let the two numbers be x and y .
According to question,
∴ xy = 192...................... (1)
x + y = 28........................(2)
As we know that,
(x - y)2 = (x + y)2 - 4xy
Put the value of xy and x + y from equation (1) and (2), we will get
(x - y)2 = (28)2 - 4 x 192
⇒ (x - y)2 = 784 - 786
⇒ (x - y)2 = 16
⇒ x - y = 4......................(3)
Add the equation (2) and (3), we will get
x + y + x - y = 28 + 4
2x = 32
x = 16
Put the value of x in equation (2), we will get
16 + y = 26
y = 28 - 16
y = 12
So numbers are x = 16 and y = 12.
Smaller number is 12.

Test: Linear Equations - 2 - Question 9

The present ages of Vikas and Vishal are in the ratio 15:8. After ten years , their ages will be in the ratio 5:3. Find their present ages

Detailed Solution for Test: Linear Equations - 2 - Question 9

Let us assume the ratio factor is x.
Therefor the present ages of Vikas and Vishal be 15x years and 8x years.
After 10 years
vikas's age = 15x + 10 and Vishal' age = 8x + 10
According to question,
(15x+10)/(8x+10) = 5/3
⇒ 3(15x + 10) = 5(8x + 10)
⇒ 45x + 30 = 40x + 50
⇒ 5x =20
⇒ x = 20/5
⇒ x = 4
Therefore Present age of Vikas = 15x = 15 x 4 = 60 years
and Present age of Vishal = 8x = 8 x 4 = 32 years

Test: Linear Equations - 2 - Question 10

The sum of the ages of a father and his son is 4 times the age of the son. If the average age of the father and the son is 28 years, What is the son's age?

Detailed Solution for Test: Linear Equations - 2 - Question 10

Let us assume father age is F and his son age is S.
According to question,
The sum of the ages of father and his son is 4 times the age of the son,
F + S = 4S
F = 3S.............. (1)
Average age of father and son is 28.
(F + S)/2 = 28
F + S = 56
Put the value of F from equation from (1),
3S + S = 56
4S = 56
S = 14 years
Age of Son = 14 years.

Test: Linear Equations - 2 - Question 11

The sum of three consecutive even numbers is 14 less than one-fourth of 176. What is middle number?

Detailed Solution for Test: Linear Equations - 2 - Question 11

Let the middle number = x , then first number = x - 2 and last number = x + 2
(x - 2) + x + (x + 2) = 176/4 - 14
x - 2 + x + x + 2 = 44 - 14
3x = 30
x = 10
So middle number = x = 10

Test: Linear Equations - 2 - Question 12

When 20 is subtracted from a number , it reduces to seven- twelve of the number. What is the sum of the digit of the number?

Detailed Solution for Test: Linear Equations - 2 - Question 12

Let the number be x.
According to question,
x - 20 = 7x/12
⇒ x - 7x/12 = 20
⇒ (12x - 7x)/12 = 20
⇒ 5x/12 = 20
⇒ x = 20 x 12/5
⇒ x = 4 x 12
⇒ x = 48
Sum of the digits of 48 = 4 + 8 = 12

Test: Linear Equations - 2 - Question 13

The cost of 5 pens and 8 pencils is ₹31.What would be the cost of 15 pens and 24 pencils ?

Detailed Solution for Test: Linear Equations - 2 - Question 13

Cost of 5 pens + 8 pencils = ₹31
On multiplying by 3
15 pens + 24 pencils = 3 x 31 =
₹15 pens + 24 pencils = 93

Test: Linear Equations - 2 - Question 14

The sum of three numbers is 300. If the ratio between first and second be 5:9 and that between second and third be 9:11, then second number is

Detailed Solution for Test: Linear Equations - 2 - Question 14

Let us assume the ratio factor is x.
Then the three numbers be 5x , 9x and 11x respectively.
Then according to question,
5x + 9x + 11x = 300
⇒ 25x = 300
⇒ x = 12
So, the second number is 9x = 9 x 12 =108

Test: Linear Equations - 2 - Question 15

Cost of 36 pens and 42 pencils is ₹ 460/- . What is the cost of 18 pens and 21 pencils ?

Detailed Solution for Test: Linear Equations - 2 - Question 15

Let us assume 1 pen cost be ₹ x and 1 pencil cost be ₹ y.
According to question,
The Cost of 36 pens and 42 pencils is 460.
36x + 42y = 460 ...................(1)
then the cost of 18 pens and 21 pencils = 18x + 21y.
Now divide the Equation (1) by 2. We will get,
18x + 21y = 230
So the Cost of 18 pens and 21 pencils = 18x + 21y = 230

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## Quantitative Aptitude for SSC CGL

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