Test: Linear Equations - 1 - SSC CGL MCQ

# Test: Linear Equations - 1 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - Test: Linear Equations - 1

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Test: Linear Equations - 1 - Question 1

### The Fourth term of an Arithmetic Progression is 37 and the Sixth term is 12 more than the Fourth term. What is the sum of the Second and Eight terms?

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

Let us assume the first number is a and common difference is d.
According to question,
4th term of A.P = 37
a + (n - 1) x d = 37
Put the value of a , n and d, we will get,
a + (4 - 1) x d = 37
a + 3d = 37..................(1)
sixth term is 12 more than the fourth term,
6th term = 12 + 4th term
a + (n - 1) x d = 12 + 37
a + (6- 1) x d = 49
a + 5d = 49................(2)
subtract the equation (1) from (2)
a + 5d - a - 3d = 49 - 37
5d - 3d= 12
2d =12
d = 6
Put the value of d in equation (1), we will get
a + 3 x 6 = 37
a = 37 - 18
a = 19
Second term = a + (n - 1) x d = 19 + (2 - 1) x 6 = 19 + 6 = 25
Eight term = a + (n - 1) x d = 19 + (8 - 1) x 6 = 19 + 42 = 61
Sum of Second and Eight term = 25 + 61 = 86
Sum of Second and Eight term = 86

Test: Linear Equations - 1 - Question 2

### On March 1st 2016 , sherry saved ₹ 1. Everyday starting from March 2nd 2016, he save ₹1 more than the previous day . Find the first date after March 1st 2016 at the end of which his total savings will be a perfect square.

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

According to the question,
Every day adding 1 rs extra to previous day.
Let us assume after n day, total saving will become perfect square.
1 + 2 + 3 + 4 + 5 + 6 +.......................+ tn.
Apply the algebra A.P formula,
sum of total rupees after n days = n(n+1)/2
Hit and trail method, Put the value of n = 2 , 3 , 4, 5 .... and so on to get the perfect square.
If n = 2
n(n+1)/2 = 2 x 3 / 2 = 3 which is not perfect Square.
If n = 3
n(n+1)/2 = 3 x 4 / 2 = 6 which is not perfect Square.
If n = 4
n(n+1)/2 = 4 x 5 / 2 = 10 which is not perfect Square.
If n = 5
n(n+1)/2 = 5 x 6 / 2 = 15 which is not perfect Square.
If n = 6
n(n+1)/2 = 6 x 7 / 2 = 21 which is not perfect Square.
If n = 7
n(n+1)/2 = 7 x 8 / 2 = 28 which is not perfect Square.
If n = 8
n(n+1)/2 = 8 x 9 / 2 = 36 which is perfect Square.

n(n+1)/2 should be a perfect square . The first value of n when this occurs would be for n = 8. thus , on the 8th of March of the required condition would come true.

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Test: Linear Equations - 1 - Question 3

### A number 15 is divided into 3 parts which are in Arithmetic Progression (A.P) and the sum of their squares is 83. What will be the smallest number?

Detailed Solution for Test: Linear Equations - 1 - Question 3

Let us assume the second number is a and the difference between consecutive numbers is d.
According to Arithmetic progression,
First number = a - d
Second number = a
Third number = a + d
According to question,
Sum of the all three numbers = 15
a - d a + d = 15
3= 15
= 5
Again according to given question,
sum of square of the 3 numbers = 83
(a - d) 2 a 2 + (a + d) 2 = 83
apply the algebra formula
2 d 2 - 2ad + a 2 + 2 d 2 + 2ad = 83
3a 2 2d 2 = 83
Put the value of in above equation.
3 x 5 2 2d2 = 83
3 x 25 2d2 = 83
75 2d2 = 83
2d 2 = 83 - 75
2d 2 = 8
d 2 = 8/2
d 2 = 4
d = 2
Put the value of and in below equation.
First number = a - d = 5 - 2 = 3
Second number = = 5
Third number = a + d = 5 + 2 = 7
The smallest number is 3.

Test: Linear Equations - 1 - Question 4

How many 3-digits numbers are completely divisible by 6?

Detailed Solution for Test: Linear Equations - 1 - Question 4

First 3-digit number divisible by 6 is 102 and last 3-digit number divisible by 6 is 996.
Difference between two consecutive numbers divisible by 6 is 6.
So 3-digit numbers divisible by 6 are 102,108,114, ......., 996.
This is an Arithmetic Progression in which a = 102, d = 6 and l = 996.
where a = First Number , l = Last Number and d = difference of two consecutive numbers.
Let the number of terms be n. So Last term = tn
Then tn = 996
Use the formula for n terms of arithmetic progression.
∴ a + ( n - 1) x d = 996
⇒ 102 + (n - 1) x 6 = 996
⇒ 6(n - 1) = 894
⇒ (n - 1) = 149
⇒ n = 150
∴ Numbers of terms = 150

Test: Linear Equations - 1 - Question 5

A man arranges to pay off a debt of ₹3600 in 40 annual installments which form an Arithmetic Progression (A.P). When 30 of the installments are paid, he dies leaving one-third of the debt unpaid . Find the value of the first installment.

Detailed Solution for Test: Linear Equations - 1 - Question 5

Let us assume the first installment is a and difference between two consecutive installments is d.
According to question,
Sum of 40 Installments = 3600
Apply the formula Sum of first n terms in an Arithmetic Progression = Sn = n/2[ 2a + (n?1)d ]
where a = the first term, d = common difference, n = number of terms.

n/2[ 2a + (n?1)d ] = 3600
Put the value of a, n and d from question,
40/2[ 2a + (40 ?1)d ] = 3600
20[ 2a + 39d ] = 3600
[ 2a + 39d ] = 3600/20 = 180
2a + 39d = 180...........................(1)
Again according to given question,
After paying the 30 installments the unpaid amount = 1/3(total unpaid amount) = 3600 x 1/3
After paying the 30 installments the unpaid amount = 1200
So After paying the 30 installments the paid amount = 3600 - 1200 = 2400
Sum of 30 installments = 2400
30/2[ 2a + (30 ?1)d ] = 2400
[ 2a + (30 ?1)d ] = 2400 x 2/30
2a + 29d = 80 x 2 = 160
2a + 29d = 160..........................(2)
Subtract the Eq. (2) from Eq. (1), we will get
2a + 39d - 2a - 29d = 180 - 160
10d = 20
d = 2
Put the value of d in Equation (1), we will get
2a + 39 x 2 = 180
2a = 180 - 78
a = 102/2
a = 51
The value of first installment = a = 51

Test: Linear Equations - 1 - Question 6

A boy agrees to work at the rate of 1 rupees on the first day, 2 rupees on the second day. Four rupees on the third day and so on. How much will the boy get if he starts working on the 1st of February and finishes on the 20th of February ?

Detailed Solution for Test: Linear Equations - 1 - Question 6

According to given question,
First day 1 rs, second day 2 rs, third day 4 rs.....
1 , 2 , 4............ and so on
The given series is in geometric progression (GP).
1, 21, 22,.............
Sum of first n terms in a geometric progression(GP)
Sn = a (rn ? 1) /(r ?1)
if r > 1 and where a= the first term, r = common ratio,n = number of terms

As per the given question a = 1 , r = 2 and n = 20;
Sn = 1(220 ? 1) /(2 ?1)
Sn = (220 ? 1) /1
Sn = (220 ? 1)
Sn = 220 ? 1

Test: Linear Equations - 1 - Question 7

If the mth term of an Arithmetic Progression (AP) is 1/n and nth term is 1/m, then find the sum of mn terms.

Detailed Solution for Test: Linear Equations - 1 - Question 7

let us assume a be the first term and d the common difference.
According to Question,
mth term of an Arithmetic Progression (AP) is 1/n
Use the Arithmetic Progression formula for mth term,
⇒ a + ( m - 1 ) x d = tm
⇒ a + ( m - 1 ) x d = 1/n
⇒ a + md - d = 1/n .......................(1)
Use the Arithmetic Progression formula for nth term,
⇒ a + ( n - 1 ) x d = tn
⇒ a + ( n - 1 ) x d = 1/m
⇒ a + nd - d = 1/m.........................(2)
Subtracts Equation (1) from Equation (2) , we will get
⇒ a + nd - d - (a + md - d) = 1/m - 1/n
⇒ a + nd - d - (a + md - d) = 1/m - 1/n
⇒ a + nd - d - a - md + d = 1/m - 1/n
⇒ nd - md = 1/m - 1/n
⇒ d(n - m) = 1/m - 1/n
⇒ d(n - m) = (n - m)/mn
⇒ d = 1/mn.......................................(3)
Put the value of d in Equation (1), we will get,
⇒ a + md - d = 1/n
⇒ a + ( m x 1/mn ) - 1/mn = 1/n
⇒ a + 1/n - 1/mn = 1/n
⇒ a = 1/n - 1/n + 1/mn
⇒ a = 1/mn.....................................(4)
Now according to question,
Sum of mn terms = mn/2(2a + ( mn - 1) x d)
Smn = mn/2(2a + ( mn - 1) x d)
Put the value of a and d in above equation, we will get,
Smn = mn/2 [ 2 x 1/mn + ( mn - 1) x 1/mn ]
Smn = mn/2 [2/mn + 1 - 1/mn ]
Smn = mn/2 [ 1 + 1/mn ]
Smn = mn/2 [ ( mn + 1 )/mn ]
Smn = mn/2 (1 + mn)/mn
Smn = 1/2 (1 + mn)
Smn = (1 + mn)/2 = (mn + 1)/2

Test: Linear Equations - 1 - Question 8

If 24 is subtracted from a number, it reduces to its four-seventh. what is the sum of the digits of that number ?

Detailed Solution for Test: Linear Equations - 1 - Question 8

Let us assume the number is n.
According to question,
If 24 is subtracted, it becomes 4/7 of the number.
n - 24 = n x 4/7
⇒ n - 4n/7 = 24
⇒ (7n - 4n)/7 = 24
⇒ 3n/7 = 24
⇒ 3n = 24 x 7
⇒ n = 24 x 7/3
⇒ n = 8 x 7
⇒ n = 56
Sum of the digits of the number = 5 + 6 = 11

Test: Linear Equations - 1 - Question 9

What is the sum of all the two-digits numbers which when divided by 7 gives a remainder of 3 ?

Detailed Solution for Test: Linear Equations - 1 - Question 9

First Two Digit Number, which is divided by 7 and give the remainder 3, will be 10.
Last Two Digit Number, which is divided by 7 and give the remainder 3, will be 94.
The common difference between two consecutive numbers will be 7.
This series will be like →10, 17, 21 ,............................ 94.
Here First Number a = 10, Last number tn = 94, Common Difference d = 7 ;
Using the Formula for Last Number tn = a + (n - 1) x d;
⇒ a + (n - 1) x d = tn
Put the value of a , d and tn in above equation.
⇒ 10 + (n - 1) x 7 = 94
⇒ 10 + 7n - 7 = 94
⇒ 7n + 3 = 94
⇒ 7n = 94 - 3 = 91
⇒ n = 91/7
⇒ n = 13
Using the formula for the sum of Arithmetic Progression.
Sn = n/2 [ First Number + Last Number ] ;
Put the value of n , First Number and Last Number, we will get,
Sn = 13/2 [ 10 + 94 ]
Sn = 104 x 13/2
Sn = 52 x 13
Sn = 676

Test: Linear Equations - 1 - Question 10

The Sum of all terms of the arithmetic progression having 10th terms except for the 1st term, is 99, and except for the 6th term , 89. Find the 3rd terms of the progression if the sum of the 1st and the 5th term is equal to 10.

Detailed Solution for Test: Linear Equations - 1 - Question 10

Let us assume the first term is a and common difference is d.
Formula for nth term of Arithmetic Progression = a + ( n - 1 ) x d
So first term t1 = a + ( n - 1 ) x d
t1 = a + (1 - 1) x d = a + 0 x d = a
Fifth term t5 = a + ( n - 1 ) x d = a + (5 - 1) x d = a + 4d
According to question,
Sum of the first term and the fifth term = 10
t1 + t5 = 10
⇒ a + a + 4d = 10
⇒ 2a + 4d = 10
⇒ a + 2d = 5.............................(1)
Formula of sum of n terms for Arithmetic Progression,
Sn = n/2[ 2a + ( n - 1 ) x d ]
Put the value of n = 10, Since total number of terms is 10 .
⇒ S10 = 10/2[ 2a + ( 10 - 1 ) x d ]
⇒ S10 = 5[ 2a + 9 x d ] = 10a + 45d
Again According to question,
The sum of all terms of the Arithmetic Progression, except for the 1st term = 99
S10 - first term = 99
⇒ 10a + 45d - a = 99
⇒ 9a +45d = 99
⇒ a + 5d = 11.................................(2)
Subtracts Equation (1) from Equation (2), we will get,
⇒ a + 5d - a - 2d = 11 - 5
⇒ 3d = 6
⇒ d = 2
Put the value of d in equation (1), we will get
⇒ a + 2 x 2 = 5
⇒ a + 4 = 5
⇒ a = 5 - 4
⇒ a = 1
Since tn = a + ( n - 1) x d
find the third term by putting the value of a , n and d. we will get,
t3 = 1 + ( 3 - 1) x 2 = 1 + 2 x 2 = 1 + 4 = 5
So 3rd term of Arithmetic Progression is 5.

Test: Linear Equations - 1 - Question 11

If the sum of one-half and one-fifth of the number exceeds one-third of that number by 7 1/3, the number is

Detailed Solution for Test: Linear Equations - 1 - Question 11

Let us assume the number is a.
Given that 7 1/3 = 22/3
According to question,
a x 1/2 + a x 1/5 = a x 1/3 + 22/3
⇒ a/2 + a/5 = a/3 + 22/3
⇒ a/2 + a/5 - a/3 = 22/3
⇒ (15a + 6a - 10a)/30 = 22/3
⇒ (15a + 6a - 10a) = 30 x 22/3
⇒ 11a = 10 x 22
⇒ a = 10 x 2
⇒ a = 20

Test: Linear Equations - 1 - Question 12

In the certain party, there was a bowl of rice for every two guests , a bowl of juice for every three of them and a bowl of meat for every four of them. If there were all 65 bowls of food , then how many guests were there in the party ?

Detailed Solution for Test: Linear Equations - 1 - Question 12

Let the number of rice bowls be a, the number of juice bowls be b, and the number of meat bowls be c.
According to question,
a + b + c = 65........................(1)
The total number of guests = 2a
The total number of guests = 3b
The total number of guests = 4c
So the total number of guests will be same in the party.
2a = 3b = 4c..........................(2)
As per Equation (2)
b = 2a/3................................(3)
c = 2a/4 = a/2......................(4)
Now put the value of b and c from Equation (3), (4) in Equation (1),
a + 2a/3 + a/2 = 65
(6a + 4a + 3a)/6 = 65
13a = 65 x 6
a = 5 x 6 = 30
Put the value of a in equation (3) and (4) in order to get the value of b and c,
b = 2 x 30/3 = 2 x 10 = 20
c = 30/2 = 15
The Total number of Guests = 2a = 3b = 4c = 60

Test: Linear Equations - 1 - Question 13

The sum of two numbers is 25 and their difference is 13. Find their product.

Detailed Solution for Test: Linear Equations - 1 - Question 13

Let us assume the numbers are a and b.
According to question,
sum of two numbers = 25
a + b = 25......................(1)
difference of two numbers = 13
a - b = 13........................(2)
add the Equation (1) and (2)
a + b+ a - b = 25 + 13
⇒ 2a = 38
⇒ a = 19
Put the value of a in equation in (1)
19 + b = 25
⇒ b = 25 - 19
⇒ b = 6
Product of the numbers = a x b
put the value of a and b,
⇒ Product of the numbers = 19 x 6
⇒ Product of the numbers = 114

Test: Linear Equations - 1 - Question 14

A driver's income consists of his salary and tips. during one week his tips were 5/4 of his salary. what fraction of his income came from tips?

Detailed Solution for Test: Linear Equations - 1 - Question 14

Let us assume the salary of driver be ₹ R
Then Tips of week = R x 5/4
According to question,
his income during one week = Salary + Tips
⇒ Total Income during one week = R + (5R/4)
⇒ Total Income during one week = (4R + 5R)/4 = 9R/4
Required Fraction = Tips in a week/Total Income in a week
⇒ Required Fraction = 5R/4 / 9R/4
⇒ Required Fraction = 5R/4 x 4/9R
⇒ Required Fraction = 5/9

Test: Linear Equations - 1 - Question 15

Ram and Mohan are friends. Each has some money. If Ram gives ₹ 30 to Mohan, Then Mohan will have twice the money left with Ram. But if Mohan gives ₹ 10 to Ram, Then Ram will have thrice as much as is left with Mohan. How much money does each have?

Detailed Solution for Test: Linear Equations - 1 - Question 15

Let us assume Ram has R rupees and Mohan has M rupees.
According to question,
If Ram gives 30 rupees to Mohan, then
Ram has left money = R - 30 and Mohan has money = M + 30
Then Mohan will have twice the money left with Ram,
M + 30 = 2(R - 30)
M + 30 = 2R - 60
2R - M = 90................................(1)
Again According to question,
if Mohan gives 10 rupees to Ram, then
Mohan has left the money = M - 10 and Ram has the money = R + 10
Then According to question,
Ram will have thrice as much as is left with Mohan,
R + 10 = 3 (M - 10 )
⇒ R + 10 = 3M - 30
⇒ 3M - R = 10 + 30
⇒ 3M - R = 40..................................(2)
After Multiplying 2 with Equation (2) , add with the equation (1),
6M - 2R + 2R - M = 80 + 90
⇒ 6M - M = 170
⇒ 5M = 170
⇒ M = 170/5
⇒ M = 34
Put the value of M in equation (1), we will get
⇒ 2R - 34 = 90
⇒ 2R = 90 + 34
⇒ 2R = 124
⇒ R = 124/2
⇒ R = 62

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## Quantitative Aptitude for SSC CGL

314 videos|170 docs|185 tests