#Algebra
Practice Examples 1, Practice Examples 2
Q.6 www.limaglobal.com
8 years ago Albert was thrice as old as Alia. Albert is 12 years older than Alia. What is the present age of Albert?
A. 22 years
B. 26 years
C. 16 years
D. 18 years
E. 20 years
Solution;
Let us take present age of Alia as ‘a’ years
So, Albert’s present age will be a+12 years
8 years ago Alia’s age = a – 8
8 years ago Albert’s age = (a + 12 ) – 8
= a + 4 ………………Equation 1
8 years ago Albert was thrice of Alia’s age = 3(a – 8) ………………Equation 2
Solve the Equation 1 and Equation 2
a +4 = 3(a – 8)
a + 4 = 3a – 24
3a – a = 24 + 4
2a = 28
a = 28/2 = 14
Alia’s present age = 14 years, So, Albert’s present age = 14 + 12 = 26 years
The answer is ‘B’.
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Q.7 www.limaglobal.com
Jar ‘M’ is 20% larger than Jar ‘N’. There is 80 ounces of apple juice in Jar ‘N’.
Quantity A Quantity B
Ounces of apple juice added in Jar ‘M’ 15.85 Ounces
to make it full
A. If Quantity A is greater
B. If Quantity B is greater
C. If Quantity A and Quantity B are equal
D. The relationship cannot be determined from given information
Solution;
Jar M is 20% larger than that of Jar ‘N’.
If we take apple juice in Jar ‘N’ as 100%, then Apple juice in Jar ‘M’ is 120%
100 % is 80 ounces
120% will be = (120 x 80)/100
= 96 ounces
Difference; 96 – 80 = 16 Ounces.
16 Ounces more apple juice needs to be added to Jar ‘M’ to make it full.
Quantity A = 16 and Quantity B = 15.85.
If we compare Quantity A with Quantity B, Quantity A is greater than Quantity B.
The Answer is ‘A’.
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Q.8 www.limaglobal.com
In an electronic store, the cost of mobile is double than the cost of an electric iron. Albert purchased mobiles and electric irons worth of $300. If Albert has purchased 50 numbers of both the items and each electric iron cost $5, then how many electric irons does he purchase?
A. 10
B. 25
C. 30
D. 40
E. 32
Solution
Let us take total numbers of mobile purchased = m
Total numbers of electric iron purchased = n
Each electric iron cost = $5
Each mobile will cost 2 x 5 = $10
Total cost; 10 x m + 5 x n = 300 ………………….Equation 1
Total numbers of mobiles and electric irons purchased = 50
m + n = 50
n = 50 – m ……………………….Equation 2
Put the value of equation 2 in equation 1
10m +5(50-m) = 300
10 m + 250 – 5m =300
5m = 300 – 250
m = 50 /5
m = 10
Albert has purchased 10 numbers of mobile and 50 – 10 = 40 numbers of electric irons.
The answer is ‘D’.
Q.9 www.limaglobal.com
Albert invested $5000 in a bank at interest compounded half yearly, if he got $5150 after one year, what is the rate of interest at which he deposited the amount?
A. 2.97
B. 2.38
C. 2.85
D. 2.67
E. 3.01
Solution;
The formula for interest rate compounded n times per year
V = P (1+r/nx100)nt
Where V = Value after interest = $ 5150
P = Principal Amount = $ 5000
n = Number of times per year = half yearly = 2
t = time = 1 year
5150 = 5000 (1 + r/200)2×1
5150 / 5000 = (1 + r/200)2
1.03 = (1 + r/200)2
Taking square root both side
√ 1.03 = (1+r/200)
1.014 – 1 = r/200
r/200 = 0. 0.014
r = 200 x 0.014
r = 2.97
So, the annual interest rate will be @ 2.97%. The answer is ‘A’.
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Q.10 www.limaglobal.com
Albert can prepare 12 pizzas in 6 hours, while Alia can produce 20 pizzas in 5 hours, how many pizzas they can prepare together in 90 minutes?
A. 6
B. 22
C. 8
D. 11
E. 9
Answer
Rate x Time = Work
For Albert’s Rate = r1 = 12/6 = 2 pizzas per hour
Alia’s Rate = r2 = 20/5 = 4 pizzas per hour
Together they can prepare 2 + 4 = Rate of Alia and Albert = 6 pizzas in an hour
Now we have to find number of pizzas in 90 minutes;
90 min. = 90 /60 = 1.5 hours
Work = Rate x Time
= 6 x 1.5
= 9
Together Albert and Alia can prepare 9 pizzas in an hour. The answer is ‘E’.
