Sample Mathematics

1. Two numbers are in the ratio 3: 2. Cube of their difference is 125. Find the numbers.
(3x – 2x) 3 = 125
x3 = 125
x = 5
The numbers are 15 and 10.


2. A bag contains 5 paise, 10 paise and 20 paise coins in the ratio 2:4:5. Total amount in the bag is Rs. 4.50. How many coins are there in 20 paise?
T = (2 x 5) + (4 x 10) + (5 x 20)
= 150 paise
= Rs. 1.50
X = 4.50/1.50
X = 3
Quantity of 20 paise coins
= 5 x 3
= 15 coins.



3. Average age of a family of four members is 34. What is the average age of the family after 4 years?
Since every ones age is increased by four in 4 years, the average will also increase
The new average = 34 + 4 = 38


4. Average age of 5 students was 18. After adding teacher’s age the average became 20. What is the age of the teacher?
Answer = (18 x 5) ~ (20 x 6) = 30


5. Average marks scored by Sai in 11 subjects is 60. Average marks scored by her in first 6 subjects is 50 and average marks scored by her in last 6 subjects is 62. Find the mark scored by Sai in 6th subject.
Ao = 60; No = 11
A1 = 50; N1 = 6
A2 = 62; N2 = 6
Substituting the values in the formula, we
get M = 50x6 + 62x6 – 60x11
= 300 + 372 – 660
=12
Marks scored by Sai in 6th subject = 12


6. Sradha’s monthly salary is Rs. 40000. If an increment of Rs. 5000 is provided, what is the percentage increase in her salary?
From 40000 to 45000
Substituting the values in the above formula, we get
% increase = [(45000 – 40000)/40000] 100
= (5000/40000)x 100
= 12.5 % increase


7. Salary of Saran was Rs. 20000. If the salary is decreased by 10 percent and then increased by 30 percent, what is the new salary of Saran?

a = first percentage change
b = second percentage change
a = -10
b = 30
Net change = -10 + 30 + [(-10)(30)/100]
= 20 – 3 = +17% (positive value = % increase) New salary of Sai
= 20000 + (17/100)20000 = 23400


8. Price of a watch was Rs. 1000. The price is increased by 10% and then decreased by 10%. What is the price of the watch now?

a = 10
Net change in percentage = 102/100 = 1% decrease.
New price of watch = 1000 – (1/100)1000
= 1000 – 10
= 990


9. A bike worth Rs. 40000 is sold for Rs. 46000. What is the profit percentage?

S.P – Selling Price.
C.P – Cost Price.
S.P = 46000; C.P = 40000
Substitute the values in the above equation, we get
P% = [(46000-40000)/40000]x100 = 15% profit


10. A product worth 3000 is sold for 10% profit and then for 20% loss. What is the overall profit or loss percentage in the sale?

a = percentage profit or loss in the first case.
b = percentage profit or loss in the second case.
Substitute ‘+’ for profit percentage.
Substitute ‘-‘ for loss percentage.
Net change = 10 – 20 + [(10)(-20)/100]
= -12
The value obtained is negative.
This indicates that a loss is incurred in the sale.
12% loss.


11. Age of father is twice that of son at present. After 5 years, sum of their ages will be 100. What is the present age of father?
Assume age of son = S
Age of father = 2S
Age of father after 5 years = 2S + 5
Age of son after 5 years = S + 5
Given,
(2S + 5) + (S + 5) = 100
3S + 10 = 100
S = 30
Present age of father = 2S = 60


12. Two friends start from their home 300 m apart and walk towards each other at a speed of 2mps and 3mps. How long (in minutes) will they take to meet each other?

D = 300m
S1 = 2mps; S2 = 3mps
T = 3000/(2+3) = 60 seconds
Time taken in minutes = 60/60 = 1 minute.


13. A man travels from a point to other. The first 100 km he travels at a speed of 25 kmph. The next 80 km he travels at a speed of 40 kmph. The last 120 km he travels at a speed of 20 kmph. What is the average speed of his journey?

= 300/12
= 25 kmph


14. A train of length 450m is travelling at a speed of 54kmph. How long will it take to cross a pole?

Speed of train ST in m/s = 54x(5/18)
= 15mps
Length of train = LT = 450m
Time taken to cross = 450/15
= 30 seconds


15. Two trains A and B of length 200m and 300 are travelling at a speed of 54kmph and 36kmph respectively in opposite directions. How long will they take to cross each other?

L1 = 200; L2 = 300
S1 + S2 = 66 + 24 = 90 kmph = 25 mps.
Substitute the values in the above equation, we get
T = (200+300)/(25)
= 20 seconds


16. It took 40 days to build 10 buildings. How long will it take to build 50 buildings using same number of employees?
W1 = 10; W2 = 50
T1 = 40; T2 = ?
Substitute the values in the above formula, we get
10/50 = 40/T2
T2 = 200 days


17. A can do a piece of work in 40 days. What is the fraction of work done by him in 25 days?
Per day work of A = 1/40
Number of days for which A worked = 25 Work done = (1/40) x 25


18. A can do a work in 30 days, B in 40 days. A works for 9 days and the remaining work is done by B. What is the fraction of work done by B?
Remaining work for B = 1 – Work done
by A Work done by A = (1/30) x 9 = 3/10
Remaining work for B = 1 – (3/10) = 7/10


19. Pipe A can fill a tank in 4 hours and pipe B can empty the tank in 5 hours. If both pipes are opened together, how long will they take to fill a complete tank?

A = 4; B = 5
Time = AB/A-B
Substitute the values in the above equation, we get Time taken to fill the tank = (4 x 5)/(-4 +5) = 20 hours


20. A man invested Rs. 10000 at 4% per annum simple interest.

• 1. What is the interest he will get at the end of 3 years?
• 2. What is the total amount earned after 3 years?

P = Principle amount invested or borrowed
R = Rate of interest per term
N = Number of terms
P = 10000; N = 3; R = 4
Substitute the values in the above equation, we get
SI = 10000 x 3 x 4/100
= 1200
Amount = Principle + Interest
Amount = 10000 + 1200 = 11200


21. There are three different bells. The first bell rings every 4 hour, the second bell rings every 6 hours and the third bell rings every 15 hours. If all the three bells are rung at same time, how long will it take for them to ring together again at the same time?
Take LCM of 4, 6, 15
LCM (4,6,15) = 60
All the three bells will ring after 60 hours.


22. What is the 100th term of the series?

2, 7, 12, 17, 22, …
a = 2;
d = 5;
n = 100
tn = 2
+ (100 – 1)5
tn = 2
+ 495 = 497
t100 = 497


23. Find the sum of the series:

2, 7, 12, 17, 22, … 497
l - Last term of the arithmetic series.
Number of terms, n =((497-2) /5)+ 1
n = 100
Sum
= (2 + 497)(100/2)
S
= 24950


24. Bhuvan has some hens and some cows. If the total number of animal-heads are 71 and the total number of feet are 228, how many hens does Bhuvan have ?
Let Bhuwan have x hens and y cows
According to the question,
x + y = 71 ...(i)
2 x + 4y = 228 ...(ii)
Multiply equation (i) by 4 and subtract equation (ii)
from it :
4x + 4y – 2x – 4y = 284 – 228
or, 2x = 56
or, x= 56/2
=28
Number of hens = 28


25. .It was calculated that 75 men could complete a piece of work in 20 days. When work was scheduled to commence, it was found necessary to send 25 men to another project. How much longer will it take to complete the work?
Before:
One day work = 1 / 20
One man’s one day work = 1 / ( 20 * 75)
Now:
No. Of workers = 50
One day work = 50 * 1 / ( 20 * 75)
The total no. of days required to complete the work = (75 * 20) / 50
= 30 days