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## Algebra Level 2 Test 6

Congratulations - you have completed Algebra Level 2 Test 6. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%%
 Question 1
If 6 yr are subtracted from the present age of Richa and the remainder is divided by 18, then  the present age of her grandson Arth  is obtained. If Arth  is 2 yr younger to Malocha, whose age is 5 yr,  then what is the age of Richa ?
 A 84 yr B 96 yr C 48 yr D 60 yr
Question 1 Explanation:
Given that the age of Malocha = 5 yrs
Present age of Arth = 3 years
Now present age of Richa = 3 x 18 + 6 = 60
 Question 2
1 yr ago, a mother was 4 times older to her son.  After 6 yr, her age becomes more than double her son’s age by 5 yr. The present ratio of their age will be
 A 13 : 12 B 3: 1 C 11: 3 D 25: 7
Question 2 Explanation:
Let the present age of mother = p years
Let the present age of son = q years
Therefore from the question we can say that
(p – 1) = 4 (q – 1 )
p = 4q -3 …………………………………….1
and p + 56 = 2 (q + 6) + 5
p = 2q + 11 ………………………………… 2
from the above two equations we can say that
4q – 3 = 2q + 11
On solving we get the values of p and q
i.e. p = 25 years
q = 7 years
so the ratio is 25 : 7
 Question 3
25 bags of wheat each weighing 40 kg costs Rs. 2750. Find the cost of 35 bags of wheat if each bag weighs 50 kg.
 A Rs. 5812.50 B Rs. 6212.50 C Rs.  4812.50 D Rs.3812.50
Question 3 Explanation:
Given the cost of wheat = 2750
So the price of one bag would be = (2750 / 25 x 40) = 2.75
Now the price of 35 bags which is of 50 kg = {2.75 / 50 x 35} = Rs. 4812.50
 Question 4
If f (0, 1) = 1, f (1, 2) = 5 ................., f (5, 6) = 61, then what is the value of f (7, 8)?
 A 56 B 112 C 113 D 117
Question 4 Explanation:
As we can see that f (0, 1) = 1 = which is 02+ 12 = 1
Similarly = f(1,2) = 12+ 22= 5
So on: hence value of f(7,8 ) = 49+64= 113
 Question 5
A farmer has decided to build a wire fence along one straight side of his property. For this, he planned to place several fence-posts at 6 m intervals, with posts fixed at both ends of the side. After he bought the posts and wire, he found that the number of posts he had bought was 5 less than required. However, he discovered that the number of posts he had bought would be just sufficient if he spaced them 8 m apart. What is the length of the side of his property and how many posts did he buy?
 A 100 m, 15 B 100 m, 16 C 120 m, 15 D 120 m, 16
Question 5 Explanation:
Let the side of the property = p m
Let the number of the posts that farmer bought = q
So when the space is 8 m
Posts will be
= p/8 +1 = q
= and when the space between is 6 m
Posts will be = p/6 +1 –y + 5
From these both the equations
p/8 – p/6 = -5
{(3 – 4)p}/24 = -5
-p/24 = -5
p = 120m
q = 16
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