Test: Linear Equations - 2 - SSC CGL MCQ

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. 

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 

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

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

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.

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

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

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.

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

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.

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

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

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

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

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