Solved Questions on Linear Equations in One Variable
1. The numerator of a fraction is 4 more than its denominator. If 3 is subtracted from the numerator and denominator both, the fraction becomes 125. What is the original fraction?
a) 11/13
b) 13/11
c) 13/17
d) 17/13
Answer: d) 17/13
Explanation: Let denominator (D) of the original fraction be x.
The numerator of a fraction is 4 more than its denominator.
Then numerator (N) be (x + 4).
Original fraction = N/D = x + 4/x
If 3 is subtracted from the numerator and denominator both, the fraction becomes 12⁄5.
According to the question,
2. The difference between the squares of two consecutive numbers is 271. What are the numbers?
a) 115, 156
b) 125, 146
c) 135, 136
d) 145, 126
Answer: c) 135, 136
Explanation: Let two consecutive numbers be x and (x+1).
The difference between the squares of two consecutive numbers is 271.
According to the question,
(x+1)2 − x2 = 271
⇒ [(x+1)(x+1)] − x2 = 271
⇒ [x(x+1)+1(x+1)] − x2 = 271
⇒ [x2+x+x+1] − x2 = 271
⇒ [x2+2x+1] − x2 = 271
⇒ x2−x2+2x+1 = 271
⇒ 2x+1 = 271
⇒ 2x = 271 − 1
⇒ 2x = 270
⇒ x = 270/2
⇒ x = 135
Numbers are: One number = x = 135
Another number = x+1 = 135 + 1 = 136
3. What is the value of ‘a’ from the linear equation (1⁄4)(5y − 30) − (7⁄3)a = 0.25 if y = 1 + a?
a) −3
b) 3
c) −6
d) 6
Answer: c) −6
Explanation:
4. An employee works in a telecommunication company on a contract of 30 days on the condition that he will receive $150 for each day he works and he will be fined $25 for each day if he is absent. How many days did he remain absent if he received $3275 in all?
a) 5 days
b) 7 days
c) 9 days
d) 11 days
Answer: b) 7 days
Explanation: Number of days for the contract = 30 days
If an employee works a day, he will get per day = $150
If he is absent, he will be fined per day = $25
Let an employee remain absent for x days.
He worked for (30−x) days.
Total money he earned = (30−x) × 150
Total money he will be fined = x × 25
At the end of contract, he gets = $3275
According to the question,
[(30−x) × 150] – [x × 25] = 3275
⇒ 4500−150x – 25x = 3275
⇒ 4500−175x = 3275
⇒ 4500−3275 = 175x
⇒ 1225 = 175x
⇒ x = 1225/175
⇒ x = 7
He remains absent for 7 days.
5. There is a distance of 425 kilometres between two points P and Q. Two buses start simultaneously from P and Q towards each other and the distance between them after 3 hours is 173 km. If the speed of one bus is 7 km/h more than the speed of other buses, what is the speed of each bus?
a) 35.5 km/hr, 40.5 km/h
b) 35.5 km/hr, 45.5 km/h
c) 38.5 km/hr, 40.5 km/h
d) 38.5 km/hr, 45.5 km/h
Answer: d) 38.5 km/hr, 45.5 km/h
Explanation: Distance between two places A and B = 425 km
Let the speed of bus B1 be x km/h.
Then, speed of bus B2 = (x+7) km/h
After 3 hours, the distance between two buses is 173 km.
Total distance travelled by two buses after 3 hours = 425 − 173
⇒ [3 × x] + [3 × (x+7)] = 425 − 173
⇒ [3x] + [3x+21] = 252
⇒ 3x + 3x+21 = 252
⇒ 6x = 252 − 21
⇒ x = 231/6
⇒ x = 38.5 km/hr
Speed of bus B1 = x km/h = 38.5 km/hr
Speed of bus B2 = (x+7) km/h = (38.5 + 7) km/hr = 45.5 km/h
