Unitary Method MCQ Quiz - Objective Question with Answer for Unitary Method - Download Free PDF
Last updated on Mar 8, 2025
Latest Unitary Method MCQ Objective Questions
Unitary Method Question 1:
Ankit runs on a circular track of length 400 m. He makes 3 rounds daily on it. How much distance (in km) does he cover on the track in a week ?
Answer (Detailed Solution Below)
Unitary Method Question 1 Detailed Solution
Concept use:
Distance Cover in a Week: The distance covered in a day × the number of days in a week.
1km = 1000m
Calculation:
Ankit runs on a circular track of length 400 m, and he completes 3 rounds daily.
Distance covered in a day = 400 m/round × 3 rounds = 1200 m.
Distance covered in a day = 1200 m / 1000 = 1.2 km.
Since Ankit runs for 7 days a week
Distance covered in a week = 1.2 km/day × 7 days = 8.4 km.
Hence, Ankit covers a distance of 8.4 kilometers on the track in a week.
Hence, the Correct option is 2.
Unitary Method Question 2:
A person purchases one kg of tea powder from each of the four places A, B, C, D at the rate of ₹1000 per 1 kg, 2 kg, 4 kg, 5 kg. If on an average he purchased x kg of tea powder per ₹ 1000, then what is the approximate value of x?
Answer (Detailed Solution Below)
Unitary Method Question 2 Detailed Solution
Given:
A person purchases one kg of tea powder
- A's place cost of sugar 1 kg = ₹ 1000
- B's place cost of sugar 2 kg = ₹ 1000
- C's place cost of sugar 4 kg = ₹ 1000
- D's place cost of sugar 5 kg = ₹ 1000
Calculation:
Total purchase = 4 kg (A + B + C + D)
cost of 1 kg tea powder in A's place = 1000
cost of 1 kg tea powder in B's place = 1000 / 2 = 500
cost of 1 kg tea powder in C' place = 1000 / 4 = 250
cost of 1 kg tea powder in D' place = 1000 / 5 = 200
Total cost = 1000 + 500 + 250 + 200 = ₹ 1950
According to the above calculation,
At ₹ 1950 = 4 kg
At ₹ 1000 = (4 / 1950) × 1000 = 2.05 kg
∴ On average, he purchases 2.05 kg of sugar for every ₹1000 spent.
Unitary Method Question 3:
Saili received a salary of Rs.18,200 for February 2014, then how many days will she have to work to get another Rs.14,950?
Answer (Detailed Solution Below)
Unitary Method Question 3 Detailed Solution
Given:
⇒ Total Salary for February = Rs.18,200
⇒ Number of Days in February 2014 = 28 (since 2014 is not a leap year)
⇒ Desired Salary = Rs.14,950
Calculation:First, calculate the Daily Salary Rate:
⇒ Daily Salary Rate = 18,200 / 28 = Rs.650 Rs/day
Next, calculate the Required Days:
⇒ Required Days = 14,950 / 650 = 23 days
Hence, Saili will have to work for 23 days to get another Rs.14,950.
∴ The correct answer is option 3.
Unitary Method Question 4:
James took a loan of Rs. 500.00 from a local moneylender and Kabeer took the same amount from cooperative society. James paid Rs. 11.00 every day for 55 days to repay the loan. Kabeer paid Rs. 600.00 to the bank after 55 days. Both the loans were closed by 55 days. Who has paid more amount and by how much?
Answer (Detailed Solution Below)
Unitary Method Question 4 Detailed Solution
Given:
Amount of loan taken by James from a local lender = Rs. 500
Amount of loan taken by Kabir from cooperative society = Rs. 500
Amount paid by James per day = Rs 11
Duration of Loan = 55 days
Calculation:
Total amount paid by James to repay the loan = 55 days x Rs 11 = Rs 605.
The total amount paid by Kabeer to repay the load = Rs 600
The difference in the amount paid the James and Kabeer = 605 - 600 = Rs 5.
So, James paid Rs 5.00 more than that of Kabeer.
Alternate Method Difference in amount paid by James and Kabeer = Rs (55 × 11) - 600 = Rs 5
Unitary Method Question 5:
If the costs of potato, tomato and onion in Rupees are 20 per kg, 20 per 500 gm and 12.25 per 250 gm respectively, then the number of coins of 25 paise in the total cost of purchasing 2 kg potatoes, 500 gm tomatoes and 0.75 kg onions are
Answer (Detailed Solution Below)
Unitary Method Question 5 Detailed Solution
Given:
The cost of potatoes for 1 kg = Rs, 20
The cost of tomatoes for 500 gm = Rs. 20
The cost of onion for 250 gm = Rs. 12.25
Calculation:
The cost for 1 kg potatoes = Rs. 20
⇒ The total cot for 2 kg potatoes = 20 × 2 = Rs. 40
⇒ The cost of tomatoes for 500 gm = Rs. 20
The cost of onion for 250 gm = Rs. 12.25
⇒ The cost of onion for 750 gm = 12.25 × 3 = Rs. 36.75
⇒ Total cost = 40 + 20 + 36.75 = Rs. 96.75
Rs. 1 = 100 paise
No. of coins of 25 paise in Rs. 1 = 4
⇒ Rs. 96.75 = 96.75 × 100 = 9675 paisa
⇒ No. of coins = (9675 × 4)/100 = 387
∴ 387 coins of 25 paisa to be needed.
The correct option is 1 i.e. 387Top Unitary Method MCQ Objective Questions
Geetanjali purchased some beads to make a garland. She had some packs of beads which contain 10 beads and some loose beads. Each bead costs 50 paise. On the purchase of 1 pack of 10 beads, she gets a reduction of Rs. 1. She purchased beads such that the total number of beads is the largest two digit even number. If she got maximum possible discount, how much money she has paid to shopkeeper?
Answer (Detailed Solution Below)
Unitary Method Question 6 Detailed Solution
Download Solution PDFGiven:
1 pack contains 10 beads.
The value of each bead = Rs. 0.5
The largest two-digit even number = 98
Calculation:
Geetanjali purchased a total of 98 beads.
1 pack contains 10 beads.
Therefore,
9 packs contain = 9 × 10 = 90 beads
The value of each bead is Rs. 0.5.
⇒ 90 × 0.5 = Rs. 45
It is given in the question that,
On the purchase of 1 pack of 10 beads, Geetanjali gets a reduction of Rs. 1.
Similarly,
On the purchase of 9 packs of 90 beads, Geetanjali gets a reduction of Rs. 9.
⇒ Rs. 45 - Rs. 9
⇒ Rs. 36
The remaining 8 beads are loose beads.
The cost of 8 beads = 8 × 0.5 = Rs. 4
Therefore,
The money paid to the shopkeeper = Rs. 36 + Rs. 4
The money paid to the shopkeeper = Rs. 40
Hence, the correct answer is Rs. 40.
Shikha went to a brick kiln to buy bricks. The rate of the bricks was 2500 rupees per one thousand bricks. How many bricks can she buy if she has only 4000 rupees?
Answer (Detailed Solution Below)
Unitary Method Question 7 Detailed Solution
Download Solution PDFGiven:
1000 bricks can be purchased in = Rs. 2500
Calculation:
In Rs. 2500 bricks can be purchased = 1000
⇒ In Rs. 1 brick can be purchased in = 1000/2500
⇒ In Rs. 4000 bricks can be purchased = \(\frac{1000}{2500} \times 4000\) = 1600
∴ 1600 bricks can be purchased in Rs. 4000.
The correct option is 2 i.e. 1600If seven persons can build a house in 30 days, how long will it take three persons to build the same house, provided that they all work at the same rate?
Answer (Detailed Solution Below)
Unitary Method Question 8 Detailed Solution
Download Solution PDFGiven:
Seven persons can build a house = 30 days
Formula Used:
Total work = Number of people × Number of days
Calculation:
Total Work = 30 × 7 = 210 units
⇒ Number of days = 210/3 = 70 days
∴ 70 days they all work at the same rate.
The correct option is 2 i.e. 70 days
An auto rickshaw meter shows charges of Rs. 25 for the first two kilometres of journey and Rs. 8 for every subsequent kilometre travelled. Manas pays Rs. 249 as fare to travel from his house to his workplace. How far is workplace from his home?
Answer (Detailed Solution Below)
Unitary Method Question 9 Detailed Solution
Download Solution PDFGiven:
Charge for first 2 km = Rs 25
The total cost of the journey = Rs 249
Calculation:
Cost of the journey for the first 2 km = 25
Cost of the remaining journey = 249 – 25 = 224
The cost of travelling 1 km after the first 2 km = Rs 8
In Rs 8 the distance travelled = 1 km
In Rs 224 the distance travelled = (1/8) × 224 = 28 km
The total distance of the journey = 2 km + 28 km = 30 km
∴ The distance between workplace and home is 30 km
The prices of fruits per kg are given below:
Mango: Rs. 80.60
Apple: Rs. 120.50
Watermelon: Rs. 22.50
Grapes: Rs. 110.40
Sneha bought \(3\frac{1}{2}\) kg watermelon, 2 kg mangoes, \(1\frac{1}{2}\) kg apples, and\(\frac{3}{4}\) kg grapes. She gave a Rs. 2000 note to the shopkeeper. How much did she get back?
Answer (Detailed Solution Below)
Unitary Method Question 10 Detailed Solution
Download Solution PDFGiven:
The prices of fruits per kg are given below:
Mango = Rs. 80.60
Apple = Rs. 120.50
Watermelon = Rs. 22.50
Grapes = Rs. 110.40
Calculation:
1) 1 kg Watermelon = Rs. 22.50
\(3\frac{1}{2}\) kg Watermelon = 3.5 kg Watermelon
⇒ The cost of 3.5 kg Watermelon = 3.5 × 22.50
⇒ The cost of 3.5 kg Watermelon = Rs. 78.75
2) 1 kg Mango = Rs. 80.60
⇒ The cost of 2 kg mangoes = 2 × 80.60
⇒ The cost of 2 kg mangoes = Rs. 161.2
3) 1 kg Apple = Rs. 120.50
\(1\frac{1}{2}\) kg apples = 1.5 kg apples
⇒ The cost of 1.5 kg apples = 1.5 × 120.50
⇒ The cost of 1.5 kg apples = Rs. 180.75
4) 1 kg grapes = Rs. 110.40
\(3\over4\) kg grapes = 0.75 kg grapes
⇒ The cost of 0.75 kg grapes = 0.75 × 110.40
⇒ The cost of 0.75 kg grapes = Rs. 82.8
The total cost of all the fruits = The cost of 3.5 kg Watermelon + The cost of 2 kg mangoes + The cost of 1.5 kg apples + The cost of 0.75 kg grapes
⇒ The total cost of all the fruits = Rs. 78.75 + Rs. 161.2 + Rs. 180.75 + Rs. 82.8
⇒ The total cost of all the fruits = Rs. 503.5
Money given to the shopkeeper by Sneha = Rs. 2000
The money that Sneha get back = Money given to the shopkeeper - The total cost of all the fruits
⇒ The money that Sneha gets back = Rs. 2000 - Rs. 503.5
⇒ The money that Sneha gets back = Rs. 1496.5
Hence, the correct answer is Rs. 1496.5
60 litre and 750 millilitre of milk is filled in 27 bottles, each of the same size. The quantity of milk in 17 such bottles is
Answer (Detailed Solution Below)
Unitary Method Question 11 Detailed Solution
Download Solution PDFGiven:
A container contains milk = 60 liters and 750 milliliters
This milk filled = 27 bottles
Concept used:
1 liter = 1000 milliliters
Calculations:
1 liter = 1000 milliliters
60 liters and 750 millimeters = 60.75 liters
Quantity of milk in each bottle = Total liters of milk ÷ number of bottles
⇒ 60.75 ÷ 27 = 2.25 liters
The quantity of milk in 17 such bottles = 2.25 × 17 = 38.25 liters
∴ The quantity of milk in 17 bottles is 38 litres and 250 millilitre
During a school tour one among the group of 15 students was elected as the group leader. If there are total 11 such groups, how many students are going for the tour?
Answer (Detailed Solution Below)
Unitary Method Question 12 Detailed Solution
Download Solution PDFGiven:
One among the group of 15 students was elected as group leader
Total number of groups = 11
Calculation:
Here, we have number of students in each group =15
So, number of students in 11 groups = 15 × 11 = 165
Hence, 165 students are going for the tour .
The cost of 12 pens and 7 pencils is ₹233. If the cost of a pen decreases by ₹1.50 and the cost of a pencil increases by ₹1, then the cost of 2 pens and 3 pencils is ₹48. What is the original cost of 4 pens and 1 pencil?
Answer (Detailed Solution Below)
Unitary Method Question 13 Detailed Solution
Download Solution PDFGiven:
Cost of 12 pens and 7 pencils is ₹233
Cost of a pen decreases by ₹1.50
Cost of a pencil increase by ₹1
Cost of 2 pens and 3 pencils is ₹48
Calculation:
Let the cost of a pen = Rs. x.
And the cost of a pencil = Rs. y.
According to question,
Cost of 12 pens and 7 pencils is ₹233
⇒ 12x + 7y = 233 ……………………(i)
New price of pen = (x - 1.50)
New price of pencil = (y + 1)
Cost of 2 pens and 3 pencils is ₹48
⇒ 2(x - 1.50) + 3(y + 1) = 48
⇒ 2x - 3 + 3y +3 = 48
⇒ 2x + 3y = 48 ……………………(ii)
Multiply eq(ii) by 6
⇒ 12x + 18y = 288 ……………………(iii)
Subtract eq(i) from eq(iii)
⇒ 11y = 55
⇒ y = 5
Put the value of y in eq(i)
⇒ 2x + 3(5) = 48
⇒ 2x + 15 = 48
⇒ 2x = 33
⇒ x = 16.5
Cost of a pen = x = Rs. 16.5
Cost of a pencil = y = Rs. 5
The cost of 4 pens and 1 pencil is,
⇒ 4 × 16.5 + 1 × 5
⇒ 66 + 5
⇒ 71
∴ Cost of 4 pens and 1 pencil is Rs. 71.Shan wishes to pour 2 litres of milk into three bottles so that the first bottle holds \(\frac 1 2\) of the milk, the second bottle holds \(\frac 3 4\) of the rest of the milk and the third bottle holds \(\frac 1 5\) of the remaining milk. How much amount of milk will be left with Shan after filling all the three bottles?
Answer (Detailed Solution Below)
Unitary Method Question 14 Detailed Solution
Download Solution PDFGiven:
Shan wishes to pour milk = 2 L
First bottle holds = 1/2 of the milk
Second bottle holds = 3/4 of the rest of the milk
Third the bottle holds = 1/5 of the remaining milk
Calculation:
Shan wishes to pour milk = 2 L
⇒ First bottle holds = 1/2 of 2 L = 1 L
⇒Remaining milk = 2 L - 1 L = 1L
Second bottle holds = 3/4 of the rest of the milk
⇒ Second bottle holds = 3/4 of 1 L = 3/4 L
⇒ Remaining milk = 1 - 3/4 = 1/4 L
1 L = 1000 ml
1/4 L = 1000/4 = 250 ml
Third bottle holds = 1/5 of remaining milk
⇒Third bottle milk = 1/5 of 250 = 50 ml
⇒ Remaining holds milk = 250 - 50 = 200 ml
∴ 200 ml of milk will be left with Shan after filling all the three bottles.
The correct option is 1 i.e. 200 mlFlora's fishing company catches 9000 kg of fresh fish every month. The selling price of fresh fish is Rs. 30/kg and the selling price of dried fish is Rs. 120/kg, but when fresh fish dried it remain \(1 \over3 \) of its weight. If flora sells dried fish, how much more money she will get than if she sells fresh fish?
Answer (Detailed Solution Below)
Unitary Method Question 15 Detailed Solution
Download Solution PDFGiven:
Weight of fish = 9000 kg
S.P. of fresh fish = Rs. 30/kg
S.P. of dried fish = Rs. 120 /kg
When fresh fish dried it remain \(1 \over 3\) of its weight.
Formula:
Price of fish = Weight × Rate
Calculation:
Price on selling fresh fish = 9000 × 30
Price on selling fresh fish = 270000
Weight of dried fish = \({1 \over 3} × 9000\) = 3000
Price on selling dried fish = 3000 × 120
Price on selling dried fish = 360000
Difference of selling price = Selling price of dried fish - Selling price of fresh fish
Difference = 360000 - 270000
Difference = Rs. 90000
∴ If flora sells dried fish, she will get Rs. 90000 more than if she sells fresh fish.