Mathematics Mcqs | Problems on L.C.M and H.C.F

In finding the HCF of two numbers, the last divisor was 41 and the successive quotients, starting from the first, where 2, 4 and 2. The numbers are______?

Byadmin July 12, 2019August 4, 2021

In finding the HCF of two numbers, the last divisor was 41 and the successive quotients, starting from the first, where 2, 4 and 2. The numbers are______?

A. 700,400
B. 820,360
C. 800,500
D. 820,369

Mathematics Mcqs | MCQs / Q&A

A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled?

A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled?

A. 4.5 hrs
B. 5 hrs
C. 6.5 hrs
D. 7.2 hrs

Read More A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled?
Mathematics Mcqs | Percentage Mcqs

605 sweets were distributed equally among children in such a way that the number of sweets received by each child is 20 % of the total number of children. How many sweets did each child receive?

605 sweets were distributed equally among children in such a way that the number of sweets received by each child is 20 % of the total number of children. How many sweets did each child receive?

A. 11
B. 24
C. 45
D. cannot be determined
Let the total number of children be x.
Then x × 20/100 × x = 605
=> x2 = 3025
=> x = √3025 = 55
Number of sweets received by each child
= (20/100 ×55) = 11

Read More 605 sweets were distributed equally among children in such a way that the number of sweets received by each child is 20 % of the total number of children. How many sweets did each child receive?
Arithmetic Mcqs | Average Mcqs | Basic Maths Mcqs | Mathematics Mcqs

In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the rate in the remaining 40 overs to reach the target of 282 runs?

In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the rate in the remaining 40 overs to reach the target of 282 runs?

A. 6.25
B. 6.5
C. 6.75
D. 7
Explanation:
Required run rate
= [282 – (3.2 * 10)]/40 = 250/40 = 6.25

Read More In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the rate in the remaining 40 overs to reach the target of 282 runs?
Mathematics Mcqs | Simplification and Approximation

The value of 25.25 – 23.23 + 24.24 is_________?

The value of 25.25 – 23.23 + 24.24 is_________?

A. 26
B. 26.26
C. 2.2
D. -22.22

Read More The value of 25.25 – 23.23 + 24.24 is_________?
Mathematics Mcqs | Profit and Loss Mcqs

After allowing a discount of 15% on the marked price, the selling price is Rs. 6800 for an article. If it was sold at marked price, there would have been a profit of 60%. The cost price of the article is?

After allowing a discount of 15% on the marked price, the selling price is Rs. 6800 for an article. If it was sold at marked price, there would have been a profit of 60%. The cost price of the article is?

A. Rs. 6400
B. Rs. 5600
C. Rs. 5000
D. Rs. 4800
E. None of these
Explanation:
Given SP = Rs. 6800
Marked price = [SP(100)]/(100 – d%) = (6800 * 100)/(100 – 15) = Rs. 8000
If SP = Rs. 8000, profit = 60%
CP = [SP(100)]/(100 + 60) = (8000 * 100)/160 = Rs. 5000

Read More After allowing a discount of 15% on the marked price, the selling price is Rs. 6800 for an article. If it was sold at marked price, there would have been a profit of 60%. The cost price of the article is?
Arithmetic Mcqs | Mathematics Mcqs | Physics Mcqs | Time & Distance Mcqs

Two automobiles are 150 kilometers apart and traveling toward each other. One automobile is moving at 60km/h and the other is moving at 40km/h mph. In how many hours will they meet?

Two automobiles are 150 kilometers apart and traveling toward each other. One automobile is moving at 60km/h and the other is moving at 40km/h mph. In how many hours will they meet?

A. 2.5
B. 2.0
C. 1.75
D. 1.5
E. 1.25

One moves at 60km/h, so in an hour, he’ll travel 60km
The other moves at 40km/h, so in an hour, he’ll travel 40km
This is towards each other, so, in an hour, we can say that they’ll get 60+40=100km closer.
This means that they get closer at 100km/h
Given that we have the distance of 150km, and the speed of closure at 100kmh, we can now use the equation
S=D/T
Rearrange to find T (times both sides by T)
S x T = D (now divide both sides by S)
T=D/S
Time = 150km divided by 100km/h
Time = 1.5 hours
Or one hour and 30 minutes

Read More Two automobiles are 150 kilometers apart and traveling toward each other. One automobile is moving at 60km/h and the other is moving at 40km/h mph. In how many hours will they meet?